Ch 10  ·  Q–
0%
Chapter 10 Exercise Solutions

Work and Energy

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

Class 9 Science Exercise NCERT Solutions Olympiad Board Exam
21 Questions
45–70 min Ideal time
Q1 Now at
Begin Q1 ↑ Progress bar
Q1
NUMERIC3 marks
Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term 'Work'.
  • Suma is swimming in a pond.
  • A donkey is carrying a load on its back.
  • A windmill is lifting water from a well.
  • A green plant is carrying out photosynthesis.
  • An engine is pulling a train.
  • Food grains are getting dried in the sun.
  • A sailboat is moving due to wind energy.
📘 Concept & Theory
Theory / Concept
›

In Physics, work has a very specific meaning. Work is said to be done only when a force acts on an object and produces displacement in the direction of the applied force or in a direction having a component along the force.

Mathematically,

\[ W = F \times s \]

where,

  • \(W\) = Work done
  • \(F\) = Applied force
  • \(s\) = Displacement of the object in the direction of force

If the displacement is not in the direction of the applied force, then the work done depends upon the angle between force and displacement.

The general expression is

\[ W = Fs\cos\theta \]

where \(\theta\) is the angle between the applied force and displacement.

Therefore, to decide whether work is done or not, always check the following conditions:

  • A force must act on the object.
  • The object must undergo displacement.
  • The displacement should have a component along the direction of the applied force.
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify whether a force is acting.

  2. Check whether the object moves.

  3. Compare the direction of force with the direction of displacement.

  4. Conclude whether mechanical work is done.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  34 steps
  1. 1. Suma is swimming in a pond
  2. Suma pushes the water backward using her hands and legs. According to Newton's Third Law, the water pushes her forward.
  3. A force is applied on the water.
  4. Due to the reaction force, Suma moves forward.
  5. There is displacement in the direction of the force acting on Suma.
  6. Therefore, mechanical work is done.
  7. 2. A donkey is carrying a load on its back
  8. The donkey applies an upward force on the load to balance its weight.
  9. Force on the load is vertically upward.
  10. The load moves horizontally with the donkey.
  11. The angle between force and displacement is \(90^\circ\)
  12. Hence,
    \[W = Fs\cos90^\circ\]
  13. Since,
    \[\cos90^\circ=0\]
  14. Therefore,
    \[W=0\]
  15. Hence, no mechanical work is done by the donkey on the load.
  16. 3. A windmill is lifting water from a well
  17. The rotating blades of the windmill convert wind energy into mechanical energy.
  18. The windmill exerts an upward force on the water.
  19. Water moves upward.
  20. Force and displacement are in the same direction.
  21. Therefore, mechanical work is done.
  22. A green plant is carrying out photosynthesis
  23. Photosynthesis is a biochemical process in which light energy is converted into chemical energy.
  24. Although energy conversion takes place, there is no mechanical force producing displacement of an object.
  25. Therefore, no mechanical work is done in the Physics sense.
  26. An engine is pulling a train
  27. The engine exerts a pulling force on the train.
  28. A forward force acts on the train.
  29. The train moves forward.
  30. Force and displacement are in the same direction.
  31. Hence, mechanical work is done.
  32. Food grains are getting dried in the sun
  33. The Sun provides heat energy, causing evaporation of water from the grains.
  34. However, the grains themselves do not undergo displacement because of an applied mechanical force.
  35. Therefore, no mechanical work is done.
  36. A sailboat is moving due to wind energy
  37. Wind strikes the sail and exerts a force on it.
  38. Wind applies force on the sail.
  39. The boat moves forward.
  40. Force causes displacement.
  41. Hence, mechanical work is done.
✅ Conclusion
›
Mechanical work is done only when a force acting on an object produces displacement along its direction (or has a component along the direction of displacement). If there is no displacement or the displacement is perpendicular to the force, then no mechanical work is done.
🎯 Exam Significance
Exam Significance
›
  • This question develops the fundamental concept of mechanical work, which is one of the most frequently tested concepts in CBSE examinations.
  • Students learn to distinguish between the everyday meaning of work and its scientific definition.
  • Questions based on identifying whether work is positive, negative or zero frequently appear in school examinations, Olympiads and entrance examinations.
  • Understanding the relationship between force and displacement forms the foundation for later topics such as energy, power, kinetic energy, potential energy and the Work-Energy Theorem.
  • The example of the donkey carrying a load is one of the most commonly asked conceptual questions in CBSE, NTSE, Olympiad and other competitive examinations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Work is done only when a force produces displacement.

  2. Both force and displacement are necessary for mechanical work.

  3. If displacement is zero, work done is zero.

  4. If force is perpendicular to displacement, then
    \[ W=0 \]

  5. Heat, light and chemical reactions may involve energy transfer, but they do not necessarily involve mechanical work.

  6. Always analyse three things: force, displacement and their relative direction before deciding whether work is done.

↑ Top
1 / 21  ·  5%
Q2 →
Q2
NUMERIC3 marks
An object thrown at a certain angle to the ground moves in a curved path and falls back to the ground. The initial and the final points of the path of the object lie on the same horizontal line. What is the work done by the force of gravity on the object?
📘 Concept & Theory
Theory / Concept
›

The force of gravity always acts vertically downward towards the centre of the Earth. The work done by gravity depends only on the change in the vertical position (height) of the object and not on the actual path followed.

The general expression for work done by a force is

\[ W = Fs\cos\theta \]

where

  • \(W\) = Work done
  • \(F\) = Applied force
  • \(s\) = Displacement
  • \(\theta\) = Angle between force and displacement

For gravity, a more useful expression is

\[ W_g = mg(h_i-h_f) \]

where

  • \(m\) = Mass of the object
  • \(g\) = Acceleration due to gravity
  • \(h_i\) = Initial height
  • \(h_f\) = Final height

Since gravity is a conservative force, the work done depends only on the initial and final heights, not on the curved trajectory of the projectile.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the initial and final positions of the object.

  2. Check whether there is any net change in height.

  3. Apply the work done by gravity formula.

  4. Apply the work done by gravity formula.

📊 Graph / Figure
Graph / Figure
›
Work Done by Gravity in Projectile Motion Initial Point Final Point Gravity (mg) Highest Point Initial Height = Final Height Net Work Done by Gravity = 0 J Positive work during descent exactly cancels negative work during ascent.
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. The object is projected at an angle and follows a parabolic (projectile) path.
  2. During its motion,
    • Gravity acts continuously in the downward direction.
    • The object first rises, reaches its highest point and then falls back.
    • The initial and final positions are on the same horizontal level.
  3. Let the initial height be
    \[h_i=h\]
  4. Since the object lands at the same horizontal level,
    \[h_f=h\]
  5. Therefore,
    \[h_i-h_f=h-h=0\]
  6. Apply the work done by gravity formula.
    \[ \begin{aligned} W_g &= mg(h_i-h_f)\\ &=mg(0)\\ &=0 \end{aligned} \]
  7. Thus, the work done by the gravitational force during the complete journey is
    \[\boxed{W_g=0\ \text{J}}\]
  8. Why is the Work Done Zero?
  9. While the object is moving upward, gravity acts opposite to the displacement and performs negative work.
  10. While the object is moving downward, gravity acts in the same direction as the displacement and performs positive work.
  11. Since the object returns to the same height, the magnitude of the negative work during ascent is exactly equal to the magnitude of the positive work during descent.
  12. Therefore,
    \[ \text{Net Work by Gravity} = \text{Positive Work} + \text{Negative Work} = 0 \]
🎯 Exam Significance
Exam Significance
›
  • This question tests the concept that gravity is a conservative force, and its work depends only on the change in height.
  • Students should understand that the shape or length of the path does not affect the work done by gravity.
  • Similar conceptual questions are frequently asked in CBSE Board examinations, NTSE, Olympiads, JEE Foundation and other school-level competitive examinations.
  • This concept is the basis for understanding gravitational potential energy, conservation of mechanical energy and the Work-Energy Theorem in higher classes.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Gravity is a conservative force.

  2. Work done by gravity depends only on the initial and final heights.

  3. The curved path of a projectile does not affect the work done by gravity.

  4. If the initial and final heights are equal,
    \[ W_g=0 \]

  5. Gravity does negative work during upward motion and positive work during downward motion.

  6. Equal positive and negative work results in zero net work when the object returns to its original height.

← Q1
2 / 21  ·  10%
Q3 →
Q3
NUMERIC3 marks
A battery lights a bulb. Describe the energy changes involved in the process.
📘 Concept & Theory
Theory / Concept
›

Energy transformation is the process in which one form of energy is converted into another. According to the Law of Conservation of Energy, energy can neither be created nor destroyed; it can only be transformed from one form to another.

A battery stores chemical energy. When it is connected to a bulb through conducting wires, chemical reactions inside the battery create a potential difference, causing electric current to flow through the circuit.

As the electric current passes through the filament of the bulb, the filament offers resistance to the flow of current. Due to this resistance, the filament becomes extremely hot and emits light. Thus, electrical energy is converted into both light energy and heat energy.

Although the primary purpose of the bulb is to produce light, a part of the electrical energy is always converted into heat. Therefore, no practical bulb is 100% efficient in producing only light.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the initial form of energy stored in the battery.

  2. Determine how this energy is converted when the circuit is completed.

  3. Identify the energy transformations occurring inside the bulb.

  4. State the complete sequence of energy conversions.

📊 Graph / Figure
Graph / Figure
›
ENERGY TRANSFORMATION IN A CIRCUIT Battery Chemical Energy Electrical Energy Bulb Light Energy Heat Energy Chemical Energy ➔ Electrical Energy ➔ Light Energy + Heat Energy
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. When the battery is connected to the bulb, a series of energy transformations takes place.
  2. The battery stores chemical energy
  3. When the circuit is completed, chemical reactions inside the battery convert the chemical energy into electrical energy.
  4. The electrical energy is carried by electric current through the connecting wires to the bulb.
  5. The filament of the bulb has high electrical resistance. As current passes through it, the filament becomes very hot due to the heating effect of electric current.
  6. The hot filament emits visible light, producing illumination. At the same time, some electrical energy is also converted into heat.
  7. Therefore, the sequence of energy transformations is
    \[ \boxed{ \text{Chemical Energy} \rightarrow \text{Electrical Energy} \rightarrow \text{Light Energy} + \text{Heat Energy} } \]
  8. Since energy is only transformed from one form to another and is neither created nor destroyed, this process obeys the Law of Conservation of Energy.
🎯 Exam Significance
Exam Significance
›
  • This question develops the concept of energy transformation, one of the fundamental topics in Physics.
  • Students should remember the correct sequence of energy conversion in an electric bulb: Chemical → Electrical → Light + Heat.
  • Questions on identifying energy transformations are frequently asked in CBSE Board examinations, Olympiads, NTSE and other school-level competitive examinations.
  • This concept forms the basis for understanding electrical appliances, electric power, efficiency and energy conservation in higher classes.
  • Competitive examinations often ask students to identify the initial, intermediate and final forms of energy involved in everyday devices.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. A battery stores chemical energy.

  2. Chemical energy is converted into electrical energy when the circuit is completed.

  3. Electrical energy is converted into light energy and heat energy inside the bulb.

  4. The filament glows because it becomes extremely hot due to its electrical resistance.

  5. The process follows the Law of Conservation of Energy.

  6. Every electrical appliance involves one or more forms of energy transformation.

← Q2
3 / 21  ·  14%
Q4 →
Q4
NUMERIC3 marks
A certain force acting on a 20 kg mass changes its velocity from 5 m s–1 to 2 m s–1. Calculate the work done by the force.
📘 Concept & Theory
Theory / Concept
›

Whenever a force changes the speed of an object, it changes the object's kinetic energy. According to the Work-Energy Theorem, the work done by the net force acting on an object is equal to the change in its kinetic energy.

Mathematically,

\[ W=\Delta K \]

Since kinetic energy is given by

\[ K=\frac{1}{2}mv^2 \]

Therefore,

\[ W=\frac{1}{2}mv^2-\frac{1}{2}mu^2 \]

or

\[ W=\frac{1}{2}m\left(v^2-u^2\right) \]

where

  • \(m\) = Mass of the object
  • \(u\) = Initial velocity
  • \(v\) = Final velocity
  • \(W\) = Work done by the force

Since the speed of the object decreases from \(5\,\text{m s}^{-1}\) to \(2\,\text{m s}^{-1}\), its kinetic energy decreases. Hence, the work done by the force will be negative.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given values.

  2. Apply the Work-Energy Theorem.

  3. Substitute the given values into the formula.

  4. Simplify step by step.

  5. State the final answer with the correct sign and SI unit.

📊 Graph / Figure
Graph / Figure
›
WORK-ENERGY THEOREM Initial State u = 5 m·s−1 Force Opposes Motion Final State v = 2 m·s−1 Work Done W = −210 J Kinetic Energy Decreases Work Done = Change in Kinetic Energy W = K2 − K1
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given
  2. Mass,
    \[m=20\,\text{kg}\]
  3. Initial velocity,
    \[u=5\,\text{m s}^{-1}\]
  4. Final velocity,
    \[v=2\,\text{m s}^{-1}\]
  5. Step-by-step Solution
  6. Using the Work-Energy Theorem,
    \[W=\frac{1}{2}m\left(v^2-u^2\right)\]
  7. Substituting the given values,
    \[ \begin{aligned} W &=\frac{1}{2}\times20\times(2^2-5^2) \\ &=10\times(4-25) \\ &=10\times(-21) \\ &=-210\ \text{J} \end{aligned} \]
  8. Therefore, the work done by the force is
    \[\boxed{W=-210\ \text{J}}\]
  9. The negative sign indicates that the force acts opposite to the direction of motion, reducing the object's kinetic energy and slowing it down.
  10. Alternative (Verification)
  11. Initial kinetic energy:
    \[ \begin{aligned} K_i &=\frac{1}{2}mu^2 \\ &=\frac{1}{2}\times20\times25 \\ &=250\ \text{J} \end{aligned} \]
  12. Final kinetic energy:
    \[ \begin{aligned} K_f &=\frac{1}{2}mv^2 \\ &=\frac{1}{2}\times20\times4 \\ &=40\ \text{J} \end{aligned} \]
  13. Hence,
    \[ \begin{aligned} W &=K_f-K_i \\ &=40-250 \\ &=-210\ \text{J} \end{aligned} \]
  14. This confirms the calculated answer.
🎯 Exam Significance
Exam Significance
›
  • This problem tests the application of the Work-Energy Theorem, one of the most important concepts in mechanics.
  • Students should understand that work done by a force can be positive, negative or zero, depending on how the force affects the object's motion.
  • The negative value of work indicates that the force opposes the motion and decreases the object's kinetic energy.
  • Numerical problems based on kinetic energy and work-energy theorem are frequently asked in CBSE Board examinations, NTSE, Olympiads and JEE Foundation-level examinations.
  • Mastering this concept helps in understanding topics such as conservation of energy, collisions and motion in higher classes.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Work done by the net force equals the change in kinetic energy.

  2. Use
    \[ W=\frac{1}{2}m(v^2-u^2) \]
    whenever the mass and velocities are known.

  3. If the final speed is less than the initial speed, the work done is negative.

  4. A negative work done indicates that kinetic energy decreases.

  5. Always write the SI unit of work as joule (J).

← Q3
4 / 21  ·  19%
Q5 →
Q5
NUMERIC3 marks
A mass of 10 kg is at a point A on a table. It is moved to a point B. If the line joining A and B is horizontal, what is the work done on the object by the gravitational force? Explain your answer.
📘 Concept & Theory
Theory / Concept
›

The work done by a force depends on the magnitude of the force, the displacement of the object, and the angle between the force and the displacement. It is given by

\[ W = Fs\cos\theta \]

where

  • \(W\) = Work done
  • \(F\) = Applied force
  • \(s\) = Displacement
  • \(\theta\) = Angle between the force and the displacement

The gravitational force always acts vertically downward towards the centre of the Earth. If an object moves only in the horizontal direction, there is no displacement in the direction of gravity. Therefore, gravity does not perform any work on the object.

This is an important application of the fact that only the component of displacement along the direction of the applied force contributes to work done.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the direction of the gravitational force.

  2. Identify the direction of displacement.

  3. Determine the angle between force and displacement.

  4. Apply the work done formula.

  5. State the final answer with justification.

📊 Graph / Figure
Graph / Figure
›
WORK DONE BY GRAVITY DURING HORIZONTAL MOTION 90° Horizontal Displacement (s) Gravity (F = mg) Position A Position B Force is Perpendicular to Displacement (F ⊥ s) W = Fs cos 90° = 0 J
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Given
    • Mass of the object,
      \[ m=10\,\text{kg} \]
    • The object moves from point A to point B along a horizontal path.
  3. Step-by-step Solution
  4. The gravitational force acting on the object is
    \[F=mg\]
  5. This force always acts vertically downward.
  6. The displacement of the object is completely horizontal.
  7. Therefore, the angle between the gravitational force and the displacement is
    \[\theta=90^\circ\]
  8. Apply the formula for work done.
    \[W=Fs\cos\theta\]
  9. Substituting
    \[\theta=90^\circ\]
  10. gives
    \[ \begin{aligned} W &=Fs\cos90^\circ \\ &=Fs\times0 \\ &=0 \end{aligned} \]
  11. Therefore, the work done by the gravitational force is
    \[\boxed{W=0\ \text{J}}\]
📘 Explanation
›

Although gravity continuously acts on the object, it does not cause any displacement in its own direction. Since the displacement is perpendicular to the gravitational force, gravity performs no mechanical work.

As the height of the object remains unchanged, its gravitational potential energy also remains unchanged.

💡 Concept Check
›
It is important to note that the mass of the object (10 kg) is not required for the calculation because the angle between force and displacement alone makes the work done equal to zero. This is a common conceptual question in examinations.
🎯 Exam Significance
Exam Significance
›
  • This question tests the understanding that work depends on the direction of force as well as the magnitude of force.
  • Students should remember that whenever force and displacement are perpendicular,
    \[ W=0. \]
  • This is one of the most frequently asked conceptual questions in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
  • The concept is fundamental for understanding gravitational potential energy, circular motion and conservative forces in higher classes.
  • Many multiple-choice questions test whether students incorrectly assume that gravity always does work simply because it acts on the object.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Gravity always acts vertically downward.

  2. The displacement of the object is horizontal.

  3. The angle between force and displacement is

    \[ 90^\circ. \]

  4. Since

    \[ \cos90^\circ=0, \]
    the work done by gravity is zero.

  5. Work depends on the direction of force and displacement, not merely on the presence of a force.

  6. Horizontal motion at a constant height does not change gravitational potential energy.

← Q4
5 / 21  ·  24%
Q6 →
Q6
NUMERIC3 marks
The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?
📘 Concept & Theory
Theory / Concept
›

Potential energy (PE) is the energy possessed by an object due to its position. For an object at a height \(h\) above the ground, the gravitational potential energy is given by

\[ PE=mgh \]

where

  • \(m\) = Mass of the object
  • \(g\) = Acceleration due to gravity
  • \(h\) = Height above the ground

Kinetic energy (KE) is the energy possessed by an object due to its motion and is given by

\[ KE=\frac{1}{2}mv^2 \]

According to the Law of Conservation of Energy, energy can neither be created nor destroyed. It can only be transformed from one form to another. During free fall, gravitational potential energy is continuously converted into kinetic energy.

Therefore, although the potential energy decreases, the total mechanical energy of the object remains constant (ignoring air resistance).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify what happens to the object's height during free fall.

  2. Determine how the potential energy changes.

  3. Determine how the kinetic energy changes.

  4. Apply the Law of Conservation of Energy.

  5. State the conclusion.

📊 Graph / Figure
Graph / Figure
›
CONSERVATION OF ENERGY DURING FREE FALL GROUND LEVEL Direction of Fall PE = Max KE = 0 State A (Rest) PE ▼ KE ▲ State B (Motion) PE = Min KE = Max State C (Impact) Mechanical Energy PE + KE = Constant Loss of Potential Energy (PE) = Gain of Kinetic Energy (KE) Energy is transformed, never created or destroyed.
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. As a freely falling object moves towards the Earth, its height above the ground decreases continuously.
  2. Since the height decreases, the gravitational potential energy also decreases because
    \[PE=mgh\]
  3. The force of gravity accelerates the object downward, increasing its speed.
  4. As the speed increases, the kinetic energy also increases according to
    \[KE=\frac{1}{2}mv^2\]
  5. The decrease in potential energy is exactly equal to the increase in kinetic energy, provided air resistance is neglected.
  6. Hence,
    \[ \text{Loss in Potential Energy} = \text{Gain in Kinetic Energy} \]
  7. Therefore, the total mechanical energy remains constant.
    \[ \begin{aligned} \text{Mechanical Energy} &=PE+KE\\ &=\text{Constant} \end{aligned} \]
  8. Thus, the decrease in potential energy does not violate the Law of Conservation of Energy. Instead, it demonstrates the transformation of one form of energy into another.
🎯 Exam Significance
Exam Significance
›
  • This question tests the understanding of the Law of Conservation of Energy, one of the most fundamental principles in Physics.
  • Students should clearly understand that energy is transformed, not destroyed.
  • Questions based on the conversion of potential energy into kinetic energy are frequently asked in CBSE Board examinations, NTSE, Olympiads and JEE Foundation-level examinations.
  • This concept forms the basis for studying mechanical energy, projectile motion, pendulums, roller coasters and satellite motion in higher classes.
  • Many objective questions test whether students know that the total mechanical energy remains constant during free fall when air resistance is neglected.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Gravitational potential energy decreases as the object falls.

  2. Kinetic energy increases because the object's speed increases.

  3. The loss of potential energy equals the gain in kinetic energy.

  4. Total mechanical energy remains constant during free fall (ignoring air resistance).

  5. Energy is transformed from one form to another; it is never created or destroyed.

  6. The free fall of an object is a classic demonstration of the Law of Conservation of Energy.
← Q5
6 / 21  ·  29%
Q7 →
Q7
NUMERIC3 marks
What are the various energy transformations that occur when you are riding a bicycle?
📘 Concept & Theory
Theory / Concept
›

In everyday life, energy continuously changes from one form to another. Riding a bicycle is an excellent example of multiple energy transformations. The energy required to pedal the bicycle comes from the food we eat. During cycling, this stored energy is converted into several other forms before finally producing motion.

According to the Law of Conservation of Energy, energy cannot be created or destroyed; it only changes from one form to another. Thus, while riding a bicycle, the total energy remains conserved, although a part of it is transformed into less useful forms such as heat and sound due to friction.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the original source of energy.

  2. Determine how muscles use this energy.

  3. Explain how pedalling transfers energy to the bicycle.

  4. Identify the useful and non-useful forms of energy produced.

  5. Write the complete sequence of energy transformations.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. While riding a bicycle, several energy transformations occur one after another.
  2. The food we eat stores chemical energy in our body.
  3. During cycling, our muscles convert this chemical energy into muscular energy, enabling our legs to pedal.
  4. The muscular energy rotates the pedals, crank, chain and wheels. Thus, muscular energy is converted into mechanical energy.
  5. As the wheels rotate, the bicycle moves forward. Therefore, mechanical energy is converted into kinetic energy of the moving bicycle and rider.
  6. During the ride, friction between the tyres and the road, friction in the chain and bearings, and air resistance convert part of the mechanical energy into heat energy. A small amount of energy is also produced as sound energy.
  7. Therefore, the sequence of energy transformations is
    \[ \boxed{ \text{Chemical Energy} \rightarrow \text{Muscular Energy} \rightarrow \text{Mechanical Energy} \rightarrow \text{Kinetic Energy} + \text{Heat Energy} + \text{Sound Energy} } \]
  8. The useful output is the kinetic energy that moves the bicycle forward, while heat and sound are unavoidable by-products caused by friction.
🎯 Exam Significance
Exam Significance
›
  • This is a standard CBSE conceptual question that tests the understanding of energy transformation.
  • Students should remember that everyday activities involve multiple energy conversions rather than a single conversion.
  • Questions based on identifying the sequence of energy transformations frequently appear in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
  • The concept helps in understanding the working of vehicles, electric motors, generators and machines in higher classes.
  • Competitive examinations often ask students to identify both the useful energy output and the energy lost due to friction.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Food is the primary source of chemical energy for cycling.

  2. Chemical energy is converted into muscular energy.

  3. Muscular energy drives the bicycle by producing mechanical energy.

  4. Mechanical energy is converted into kinetic energy of the moving bicycle.

  5. Some energy is always converted into heat and sound because of friction.

  6. The total energy remains conserved throughout the process.

← Q6
7 / 21  ·  33%
Q8 →
Q8
NUMERIC3 marks
Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?
📘 Concept & Theory
Theory / Concept
›

According to Physics, work is done only when a force produces displacement in the direction of the applied force. If there is no displacement, then no mechanical work is done on the object, regardless of how large the applied force is.

The work done by a force is given by

\[ W = Fs\cos\theta \]

where

  • \(W\) = Work done
  • \(F\) = Applied force
  • \(s\) = Displacement of the object
  • \(\theta\) = Angle between force and displacement

If the displacement is zero,

\[ s=0 \]

Therefore,

\[ \begin{aligned} W &= Fs\cos\theta\\ &=F\times0\times\cos\theta\\ &=0 \end{aligned} \]

Although no mechanical work is done on the object, the person pushing the object still expends energy. The body's muscles continuously consume chemical energy obtained from food to produce force.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Determine whether the rock moves.

  2. Apply the condition for mechanical work.

  3. Explain whether energy is transferred to the rock.

  4. Describe how the energy supplied by the body is utilized.

  5. State the conclusion.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. When we push a huge rock with all our strength but it does not move, the displacement of the rock is zero.
  2. A force is applied to the rock.
  3. The rock does not move. Therefore,
    \[s=0\]
  4. Using the work done formula,
    \[ \begin{aligned} W &=Fs\cos\theta\\ &=F\times0\times\cos\theta\\ &=0 \end{aligned} \]
  5. Hence, no mechanical work is done on the rock because there is no displacement.
  6. Although the rock does not gain mechanical energy, the muscles of our body continue to consume chemical energy while trying to push it.
  7. This chemical energy is mainly converted into
    • Heat energy in the muscles and surrounding tissues.
    • A small amount of sound energy due to muscular activity and contact with the rock.
    • Internal energy required for muscle contraction and other physiological processes.
  8. This is why we feel tired and our body becomes warm after pushing a heavy object even though it has not moved.
  9. herefore, the energy spent is not transferred to the rock as mechanical work; instead, it is transformed mainly into heat energy within our body.
🎯 Exam Significance
Exam Significance
›
  • This is one of the most frequently asked conceptual questions in the CBSE Class 9 examination.
  • It helps students distinguish between applying a force and doing mechanical work.
  • Students should remember that force alone is not sufficient for work; displacement is equally essential.
  • Similar questions appear in Olympiads, NTSE, JEE Foundation and other competitive examinations to test conceptual understanding.
  • The concept is important for understanding energy transfer, muscular work, efficiency and the Work-Energy Theorem in higher classes.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Mechanical work requires both force and displacement.

  2. If displacement is zero, the work done on the object is zero.

  3. No mechanical energy is transferred to the rock if it does not move.

  4. The body's chemical energy is converted mainly into heat during muscular effort.

  5. Feeling tired after pushing a heavy object is evidence that your body has expended energy even though no mechanical work was done on the object.

  6. Energy is conserved because it is transformed into other forms rather than being destroyed.

← Q7
8 / 21  ·  38%
Q9 →
Q9
NUMERIC3 marks
A certain household has consumed 250 units of energy during a month. How much energy is this in joules?
📘 Concept & Theory
Theory / Concept
›

Electrical energy consumed in homes is measured in kilowatt-hour (kWh), commonly known as a unit of electricity. The electricity bill records the energy consumed in units rather than in joules because the joule is a very small unit for everyday electrical energy consumption.

The relationship between unit of electricity and joule is

\[ 1\ \text{unit} = 1\ \text{kWh} \]

Since

\[ 1\ \text{kW}=1000\ \text{W} \]

and

\[ 1\ \text{hour}=60\times60=3600\ \text{s}, \]

therefore,

\[ \begin{aligned} 1\ \text{kWh} &=1000\times3600\ \text{J}\ &=3.6\times10^6\ \text{J} \end{aligned} \]

This conversion factor is frequently used in numerical problems involving electrical energy.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given energy consumption in units.

  2. Convert units into kilowatt-hours.

  3. Use the conversion \(1\ \text{kWh}=3.6\times10^6\ \text{J}\).

  4. Multiply to obtain the energy in joules.

  5. Write the answer in scientific notation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. Given
  2. Total electrical energy consumed
    \[=250\ \text{units}\]
  3. Step-by-step Solution
  4. Since
    \[1\ \text{unit}=1\ \text{kWh}\]
  5. therefore,
    \[250\ \text{units}=250\ \text{kWh}\]
  6. Also,
    \[ \begin{aligned} 1\ \text{kWh} &=1000\times3600\ \text{J}\\ &=3.6\times10^6\ \text{J} \end{aligned} \]
  7. Hence,
    \[ \begin{aligned} 250\ \text{kWh} &=250\times3.6\times10^6\ \text{J}\\ &=900\times10^6\ \text{J}\\ &=9\times10^8\ \text{J} \end{aligned} \]
  8. Therefore, the energy consumed is
    \[ \boxed{9\times10^8\ \text{J}}\]
🎯 Exam Significance
Exam Significance
›
  • This question tests the understanding of the commercial unit of electrical energy (kilowatt-hour).
  • Students should memorise the important conversion:
    \[ 1\ \text{kWh}=3.6\times10^6\ \text{J} \]
  • Numerical problems based on unit conversion are frequently asked in CBSE Board examinations, NTSE, Olympiads and JEE Foundation-level examinations.
  • This concept is useful while studying electric power, electricity consumption and household electricity bills in higher classes.
  • Competitive examinations often test the ability to convert between commercial and SI units of energy quickly and accurately.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. One unit of electrical energy is equal to one kilowatt-hour.

  2. The SI unit of energy is the joule (J).

  3. The standard conversion is

    \[ 1\ \text{kWh}=3.6\times10^6\ \text{J}. \]

  4. Always convert units into kilowatt-hours before converting to joules.

  5. Express large answers in scientific notation whenever possible.

← Q8
9 / 21  ·  43%
Q10 →
Q10
NUMERIC2 marks
An object of mass 40 kg is raised to a height of 5 m above the ground. What is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half-way down.
📘 Concept & Theory
Theory / Concept
›

An object raised above the ground possesses gravitational potential energy due to its position. The gravitational potential energy is given by

\[ PE=mgh \]

where

  • \(m\) = Mass of the object
  • \(g\) = Acceleration due to gravity
  • \(h\) = Height above the ground

When the object is released, its gravitational potential energy gradually converts into kinetic energy. If air resistance is neglected, the total mechanical energy remains constant according to the Law of Conservation of Energy.

The kinetic energy of the object is

\[ KE=\frac{1}{2}mv^2 \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Calculate the initial potential energy of the object.

  2. Determine the height of the object when it is halfway down.

  3. Calculate the remaining potential energy at that position.

  4. Apply the Law of Conservation of Energy to find the kinetic energy.

  5. Verify the answer using the equation of motion and the kinetic energy formula.

📊 Graph / Figure
Graph / Figure
›
ENERGY CONVERSION DURING FREE FALL GROUND LEVEL Falling Object A 5.0 m B 2.5 m C 0 m Energy Distribution Dashboard A Top State (5.0 m) PE = 2000 J KE = 0 J B Midway (2.5 m) PE = 1000 J KE = 1000 J C Ground (0 m) PE = 0 J KE = 2000 J Total Mechanical Energy (PE + KE): 2000 J (Constant) Potential Energy ▼ = Kinetic Energy ▲
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. Given
  2. Mass of the object,
    \[m=40\,\text{kg}\]
  3. Height,
    \[h=5\,\text{m}\]
  4. Taking
    \[g=10\,\text{m s}^{-2}\]
  5. Part I: Potential Energy at a Height of 5 m
  6. Using the formula,
    \[PE=mgh\]
  7. Substituting the given values,
    \[ \begin{aligned} PE &=40\times10\times5\\ &=2000\ \text{J} \end{aligned} \]
  8. Therefore,
    \[\boxed{PE=2000\ \text{J}}\]
  9. Part II: Kinetic Energy When the Object is Half-way Down
  10. When the object has fallen halfway, it has fallen through a distance of
    \[\frac{5}{2}=2.5\,\text{m}\]
  11. Therefore, its height above the ground is
    \[5-2.5=2.5\,\text{m}\]
  12. Using the Law of Conservation of Energy
  13. Initial mechanical energy
    \[=2000\ \text{J}\]
  14. Remaining potential energy at a height of \(2.5\,\text{m}\):
  15. \[ \begin{aligned} PE &=mgh\\ &=40\times10\times2.5\\ &=1000\ \text{J} \end{aligned} \]
  16. Since
    \[\text{Mechanical Energy}=PE+KE\]
  17. we get
    \[ \begin{aligned} KE &=2000-1000\\ &=1000\ \text{J} \end{aligned} \]
  18. Therefore,
    \[\boxed{KE=1000\ \text{J}}\]
🎯 Exam Significance
Exam Significance
›
  • This numerical problem combines the concepts of gravitational potential energy, kinetic energy and the Law of Conservation of Energy.
  • Students should understand that as an object falls, its potential energy decreases while its kinetic energy increases by the same amount.
  • The solution demonstrates two valid approaches: using the conservation of energy and using the equations of motion.
  • Similar numericals are frequently asked in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
  • Mastering this concept provides a strong foundation for solving advanced problems involving mechanical energy in higher classes.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Gravitational potential energy is calculated using

    \[ PE=mgh. \]

  2. Kinetic energy is calculated using

    \[ KE=\frac{1}{2}mv^2. \]

  3. During free fall, potential energy is continuously converted into kinetic energy.

  4. In the absence of air resistance,

    \[ PE+KE=\text{Constant}. \]

  5. At half the original height in this problem, both the potential energy and kinetic energy are 1000 J.

← Q9
10 / 21  ·  48%
Q11 →
Q11
NUMERIC2 marks
What is the work done by the force of gravity on a satellite moving round the Earth? Justify your answer.
📘 Concept & Theory
Theory / Concept
›

An artificial satellite revolves around the Earth in a nearly circular orbit under the influence of the Earth's gravitational force. The gravitational force continuously acts towards the centre of the Earth and provides the necessary centripetal force required for circular motion.

The work done by a force is given by

\[ W = Fs\cos\theta \]

where

  • \(W\) = Work done
  • \(F\) = Applied force
  • \(s\) = Displacement
  • \(\theta\) = Angle between the force and the displacement

In uniform circular motion, the instantaneous displacement of the satellite is always along the tangent to the circular path, whereas the gravitational force always acts towards the centre of the Earth.

Therefore, the angle between the gravitational force and the instantaneous displacement is

\[ \theta=90^\circ \]

Since

\[ \cos90^\circ=0, \]

the work done by gravity is zero.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the direction of the gravitational force.

  2. Identify the direction of the satellite's instantaneous displacement.

  3. Determine the angle between force and displacement.

  4. Apply the work done formula.

  5. State the conclusion with justification.

📊 Graph / Figure
Graph / Figure
›
GRAVITY & MOTION OF A SATELLITE Zero Work Done in a Circular Orbit Earth 90° Gravity Force (F) Tangential Displacement (s) Satellite W = F · s · cos(90°) = 0 Joules Since Force is perpendicular to Displacement, no work is done.
The gravitational force acts towards the centre of the Earth, whereas the satellite moves tangentially to its orbit. Hence, the angle between force and displacement is \(90^\circ\).
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. The gravitational force always acts towards the centre of the Earth.
  2. The satellite moves along the tangent to its circular orbit at every instant.
  3. Therefore, the angle between the gravitational force and the instantaneous displacement is
    \[\theta=90^\circ\]
  4. Apply the formula for work done.
    \[ \begin{aligned} W &=Fs\cos\theta\\ &=Fs\cos90^\circ\\ &=Fs\times0\\ &=0 \end{aligned} \]
  5. Therefore, the work done by the gravitational force on the satellite is
    \[\boxed{W=0\ \text{J}}\]
  6. Gravity continuously changes the direction of the satellite's velocity but not its speed. Since the speed remains constant, the kinetic energy of the satellite also remains constant. Hence, gravity does not transfer mechanical energy to the satellite even though it continuously changes its direction of motion.
🎯 Exam Significance
Exam Significance
›
  • This is one of the most frequently asked conceptual questions from the chapter Work and Energy in CBSE Board examinations.
  • It tests the understanding that work depends on the angle between force and displacement, not merely on the presence of a force.
  • Students should remember that gravity acts as the centripetal force for satellites moving in circular orbits.
  • Similar questions are commonly asked in Olympiads, NTSE, JEE Foundation and other competitive examinations.
  • This concept forms the foundation for understanding circular motion, satellite dynamics, gravitational fields and orbital mechanics in higher classes.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Gravity acts towards the centre of the Earth.

  2. The satellite's instantaneous displacement is tangential to its orbit.

  3. The angle between force and displacement is

    \[ 90^\circ. \]

  4. Since

    \[ \cos90^\circ=0, \]
    the work done by gravity is zero.

  5. Gravity changes only the direction of the satellite's velocity, not its speed.

  6. No mechanical energy is transferred by gravity during uniform circular motion.

← Q10
11 / 21  ·  52%
Q12 →
Q12
NUMERIC3 marks
an there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.
📘 Concept & Theory
Theory / Concept
›

Displacement is the shortest straight-line distance between the initial and final positions of an object, measured in a specified direction. An object undergoes displacement whenever its position changes.

According to Newton's First Law of Motion (Law of Inertia), an object continues to remain at rest or move with uniform velocity in a straight line unless acted upon by an unbalanced external force.

This means that an external force is not required to keep an object moving. A force is needed only to

  • Start the motion of an object at rest.
  • Stop a moving object.
  • Change its speed.
  • Change its direction of motion.

Therefore, if an object is already moving with constant velocity and no external force acts on it, it will continue to move and hence continue to have displacement.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Recall Newton's First Law of Motion.

  2. Determine whether the object is initially at rest or already moving.

  3. Explain how displacement occurs in the absence of an external force.

  4. State the final conclusion.

📊 Graph / Figure
Graph / Figure
›
WORK DONE & UNIFORM MOTION NCERT Class 9 Science • Chapter 10: Work and Energy SMOOTH FRICTIONLESS SURFACE m Initial Position m Final Position Constant Velocity (v) Displacement (s) 1. CONDITION FOR WORK DONE Applied External Force (F) = 0 2. MATHEMATICAL CALCULATION Work Done (W) = F × s   ➔   W = 0 × s   ➔   W = 0 J (Zero Work) 💡 NCERT Fact: Even though displacement (s) exists, Work Done is zero because no external force causes the motion!
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Yes, an object can have displacement even when no external force acts on it.
  2. Consider an object that is already moving with a constant velocity on a frictionless surface.
  3. According to Newton's First Law, in the absence of any unbalanced external force, the object continues to move with the same speed in the same direction.
  4. Since the object keeps changing its position with time, it undergoes displacement.
  5. Thus, displacement is possible even when the net external force acting on the object is zero.
  6. However, if the object is initially at rest, an external force is required to set it into motion. Once it starts moving, no additional external force is needed to maintain uniform motion, provided friction and other resistive forces are absent.
  7. Therefore, the correct conclusion is:
  8. An object already in motion can continue to have displacement even in the absence of any external force.
🎯 Exam Significance
Exam Significance
›
  • This question tests the understanding of Newton's First Law of Motion and the concept of inertia.
  • Students should clearly distinguish between the conditions required to start motion and those required to maintain motion.
  • Similar conceptual questions are frequently asked in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
  • This concept forms the basis for understanding inertial frames, momentum and Newton's Laws in higher classes.
  • Competitive examinations often test the misconception that a continuous force is required to keep an object moving.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  5 points

  1. Displacement means a change in the position of an object.

  2. A force is required to change the state of motion, not to maintain uniform motion.

  3. An object already in motion continues to move with constant velocity if no unbalanced external force acts on it.

  4. Displacement is possible even when the net external force is zero.

  5. Friction and air resistance are the main reasons moving objects eventually come to rest in everyday life.

← Q11
12 / 21  ·  57%
Q13 →
Q13
NUMERIC3 marks
A person holds a bundle of hay over his head for 30 minutes and gets tired. Has he done some work or not? Justify your answer.
📘 Concept & Theory
Theory / Concept
›

In Physics, mechanical work is done only when a force acting on an object produces displacement in the direction of the applied force.

The work done by a force is given by

\[ W=Fs\cos\theta \]

where

  • \(W\) = Work done
  • \(F\) = Applied force
  • \(s\) = Displacement of the object
  • \(\theta\) = Angle between the force and displacement

If there is no displacement, then

\[ s=0 \]

Therefore,

\[ \begin{aligned} W &=Fs\cos\theta\\ &=F\times0\times\cos\theta\\ &=0 \end{aligned} \]

It is important to distinguish between the everyday meaning of work and its scientific meaning. A person may feel tired after holding an object, but this does not necessarily mean that mechanical work has been done on the object.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Determine whether a force is applied.

  2. Check whether the bundle of hay undergoes displacement.

  3. Apply the work done formula.

  4. Explain why the person still becomes tired.

  5. State the final conclusion.

📊 Graph / Figure
Graph / Figure
›
HOLDING A BUNDLE OF HAY NCERT Class 9 Science • Chapter 10: Work and Energy BUNDLE OF HAY Fapplied Fgravity (mg) s = 0 No Displacement CONCEPT EXAMINATION • Upward Force is exerted: Yes (F > 0) • Displacement of the bundle: No (s = 0) Work Done Formula: W = F × s = F × 0 = 0 J No mechanical work is done on the bundle of hay. 🤔 Why does the person feel tired? The body spends internal energy as muscles undergo micro-contractions (producing heat). However, in scientific terms, zero work is done because the displacement (s) is zero.
✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. The person exerts an upward force on the bundle of hay to balance its weight.
  2. A force is applied on the bundle.
  3. The bundle remains stationary above the person's head throughout the 30 minutes.
  4. Therefore,
    \[s=0\]
  5. Using the work done formula,
    \[ \begin{aligned} W &=Fs\cos\theta\\ &=F\times0\times\cos\theta\\ &=0 \end{aligned} \]
  6. Hence, the mechanical work done on the bundle of hay is zero.
  7. Even though no mechanical work is done on the bundle, the person's muscles remain continuously contracted to support the weight.
  8. During this process, the muscles consume chemical energy obtained from food. This energy is mainly converted into
    • Heat energy
    • Internal energy required for muscle contraction
  9. As a result, the person becomes tired even though no mechanical work is performed on the bundle.
  10. Therefore, The person does not do any mechanical work on the bundle of hay because there is no displacement, although the body expends energy internally.
  11. Energy Transformation in the Human Body
  12. \[ \boxed{ \text{Chemical Energy (Food)} \rightarrow \text{Muscular Activity} \rightarrow \text{Heat Energy} } \]
🎯 Exam Significance
Exam Significance
›
ul>
  • This is one of the most frequently asked conceptual questions in the CBSE Class 9 examination.
  • It helps students differentiate between the scientific definition of work and its everyday meaning.
  • Students should remember that both force and displacement are necessary for mechanical work.
  • Similar questions appear regularly in Olympiads, NTSE, JEE Foundation and other competitive examinations.
  • This concept forms the basis for understanding work, energy, muscle physiology and efficiency in higher classes.
    • 🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  6 points

    1. Mechanical work requires both force and displacement.

    2. If displacement is zero, the work done is zero.

    3. Holding an object stationary does not involve mechanical work on the object.

    4. The human body still consumes chemical energy to maintain muscle contraction.

    5. Feeling tired does not necessarily imply that mechanical work has been done.

    6. Always distinguish between the everyday meaning and the scientific meaning of "work."

    ← Q12
    13 / 21  ·  62%
    Q14 →
    Q14
    NUMERIC3 marks
    An electric heater is rated 1500 W. How much energy does it use in 10 hours?
    📘 Concept & Theory
    Theory / Concept
    ›

    Power is the rate at which work is done or energy is consumed. It indicates how quickly an electrical appliance converts electrical energy into other forms of energy.

    The SI unit of power is the watt (W).

    One watt means that one joule of energy is consumed every second.

    \[ 1\ \text{W}=1\ \text{J s}^{-1} \]

    The relationship between power, energy and time is

    \[ P=\frac{E}{t} \]

    Therefore,

    \[ E=P\times t \]

    where

    • \(P\) = Power (W)
    • \(E\) = Energy consumed (J)
    • \(t\) = Time (s)

    Since power is given in watts, the time must always be converted into seconds before using the formula to obtain the answer in joules.

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Write the given values.

    2. Convert the time from hours into seconds.

    3. Apply the relation \(E=P\times t\).

    4. Substitute the values.

    5. Express the answer in joules and scientific notation.

    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  6 steps
    1. Given
    2. Power of the heater,
      \[P=1500\ \text{W}\]
      Time of operation,
      \[t=10\ \text{hours}\]
    3. Step-by-step Solution
    4. Convert the time into seconds.
      \[ \begin{aligned} t &=10\times60\times60\\ &=36000\ \text{s} \end{aligned} \]
    5. Apply the formula
      \[E=P\times t\]
    6. Substitute the given values.
      \[ \begin{aligned} E &=1500\times36000\\ &=54\,000\,000\ \text{J} \end{aligned} \]
    7. Express the answer in scientific notation.
      \[ \begin{aligned} E &=54\,000\,000\ \text{J}\\ &=5.4\times10^7\ \text{J} \end{aligned} \]
    8. Therefore, the electrical energy consumed by the heater in 10 hours is
      \[\boxed{5.4\times10^7\ \text{J}}\]
    🎯 Exam Significance
    Exam Significance
    ›
    • This numerical problem tests the relationship between power, energy and time.
    • Students should remember the important formula
      \[ E=P\times t. \]
    • Always convert time into seconds when power is given in watts and the answer is required in joules.
    • Similar numerical problems are frequently asked in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
    • This concept is also useful while studying electrical appliances, household electricity consumption and electric power in higher classes.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  5 points

    1. Power is the rate of consumption of energy.

    2. The SI unit of power is the watt (W).

    3. Use
      \[ E=P\times t \]
      to calculate energy consumed.

    4. Convert hours into seconds before using SI units.

    5. An electric heater rated at 1500 W consumes \(5.4\times10^7\ \text{J}\) of energy in 10 hours.

    ← Q13
    14 / 21  ·  67%
    Q15 →
    Q15
    NUMERIC3 marks
    Illustrate the law of conservation of energy by discussing the energy changes which occur when we draw a pendulum bob to one side and allow it to oscillate. Why does the bob eventually come to rest? What happens to its energy eventually? Is it a violation of the law of conservation of energy?
    📘 Concept & Theory
    Theory / Concept
    ›

    A simple pendulum consists of a small heavy bob suspended from a fixed support by a light, inextensible string. When the bob is displaced from its mean position and released, it oscillates to and fro under the action of gravity.

    During its motion, the pendulum continuously converts gravitational potential energy into kinetic energy and vice versa.

    According to the Law of Conservation of Energy,

    Energy can neither be created nor destroyed. It can only be transformed from one form into another.

    For an ideal pendulum (neglecting air resistance and friction), the total mechanical energy remains constant throughout the motion.

    Total Mechanical Energy is

    \[ \text{Mechanical Energy} = PE+KE = \text{Constant} \]

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Describe the energy at the highest position.

    2. Explain the energy conversion while the bob moves downward.

    3. Describe the energy at the mean (lowest) position.

    4. Explain the motion towards the opposite side.

    5. Discuss why the oscillations gradually stop.

    6. Relate the observations to the Law of Conservation of Energy.
    📊 Graph / Figure
    Graph / Figure
    ›
    Law of Conservation of Energy A C B Positions A & C PE = Maximum KE = Zero (Instant Stop) Position B (Lowest Point) PE = Minimum / Zero KE = Maximum (Max Speed) Total Energy PE + KE = Constant
    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  19 steps
    1. Consider a pendulum bob pulled to one side and released from rest.
    2. 1. At the Extreme Position (A)
    3. At the highest point, the bob is momentarily at rest.
      • Velocity = 0
      • Kinetic Energy = Minimum (Zero)
      • Potential Energy = Maximum
    4. Since the bob is at the greatest height, its gravitational potential energy is maximum.
    5. 2. While Moving Towards the Mean Position (B)
    6. As the bob swings downward,
      • Its height decreases.
      • Potential energy decreases continuously.
      • Its speed increases.
      • Kinetic energy increases continuously.
    7. Thus,
      \[ \text{Potential Energy} \longrightarrow \text{Kinetic Energy} \]
    8. 3. At the Mean Position (B)
    9. At the lowest point of the swing,
      • Speed is maximum.
      • Kinetic Energy is maximum.
      • Potential Energy is minimum.
    10. At this position,
      \[KE=\text{Maximum}\]
    11. 4. While Moving Towards the Other Extreme Position (C)
    12. As the bob rises upward,
      • Speed decreases.
      • Kinetic energy decreases.
      • Potential energy increases.
    13. Thus,
      \[ \text{Kinetic Energy} \longrightarrow \text{Potential Energy} \]
    14. 5. At the Other Extreme Position (C)
    15. Once again,
      • Velocity = 0
      • Potential Energy = Maximum
      • Kinetic Energy = Minimum (Zero)
    16. Therefore, during every oscillation,
      \[ \boxed{ PE \longleftrightarrow KE } \]
    17. while
      \[ \boxed{ PE+KE=\text{Constant} } \]
    18. Why Does the Pendulum Eventually Come to Rest?
    19. In practice, a pendulum does not oscillate forever because of
      • Air resistance.
      • Friction at the point of suspension (pivot).
      • Internal friction within the string and the bob.
    20. These resistive forces gradually convert the pendulum's mechanical energy into
      • Heat energy.
      • Sound energy.
    21. Consequently, the amplitude of oscillation decreases gradually until the bob finally comes to rest.
    22. Does This Violate the Law of Conservation of Energy?
    23. No.
    24. The pendulum loses mechanical energy, but the total energy of the system and its surroundings remains conserved.
    25. The mechanical energy is simply transformed into heat and sound energy.
      \[ \boxed{ \text{Mechanical Energy} \rightarrow \text{Heat Energy} + \text{Sound Energy} } \]
    26. Therefore, the Law of Conservation of Energy is never violated.
    🎯 Exam Significance
    Exam Significance
    ›
    • This is one of the most important long-answer questions in the chapter and is frequently asked in CBSE Board examinations.
    • It tests the concepts of potential energy, kinetic energy, mechanical energy and the Law of Conservation of Energy simultaneously.
    • Students should clearly remember the energy distribution at the highest position, mean position and intermediate positions.
    • Similar conceptual and diagram-based questions frequently appear in Olympiads, NTSE, JEE Foundation and other competitive examinations.
    • This concept forms the basis for studying oscillations, harmonic motion and energy conservation in higher classes.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  6 points

    1. At the extreme positions, potential energy is maximum and kinetic energy is zero.

    2. At the mean position, kinetic energy is maximum and potential energy is minimum.

    3. During oscillation, potential energy and kinetic energy continuously transform into each other.

    4. In an ideal pendulum,

      \[ PE+KE=\text{Constant}. \]

    5. Air resistance and friction gradually convert mechanical energy into heat and sound energy.

    6. The pendulum eventually comes to rest, but the Law of Conservation of Energy is never violated.

    ← Q14
    15 / 21  ·  71%
    Q16 →
    Q16
    NUMERIC3 marks
    An object of mass \(m\) is moving with a constant velocity \(v\). How much work should be done on the object in order to bring the object to rest?
    📘 Concept & Theory
    Theory / Concept
    ›

    According to the Work-Energy Theorem, the work done by the net force acting on an object is equal to the change in its kinetic energy.

    Mathematically,

    \[ W=\Delta KE \]

    Since kinetic energy is given by

    \[ KE=\frac{1}{2}mv^2 \]

    the work done can be written as

    \[ W=KE_f-KE_i \]

    where

    • \(KE_i\) = Initial kinetic energy
    • \(KE_f\) = Final kinetic energy

    When an object is brought to rest, its final velocity becomes zero. Therefore, its kinetic energy decreases from its initial value to zero. Hence, the work done on the object is negative, indicating that energy is removed from the object.

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Find the initial kinetic energy of the object.

    2. Determine the final kinetic energy after the object comes to rest.

    3. Apply the Work-Energy Theorem.

    4. Simplify the expression.

    5. Interpret the negative sign of the answer.

    📊 Graph / Figure
    Graph / Figure
    ›
    Work-Energy Theorem Displacement (d) INITIAL STATE m KE₁ = ½mv² Initial Velocity (v) Opposing Force (F) FINAL STATE m STOPPED KE₂ = 0 Work Done = Change in Kinetic Energy (ΔKE) W = KE₂ − KE₁ = 0 − ¹/₂mv² = −¹/₂mv²
    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  11 steps
    1. Given
    2. Mass of the object,
      \[m=m\]
    3. Initial velocity,
      \[v=v\]
    4. Final velocity,
      \[v_f=0\]
    5. Step-by-step Solution
    6. Calculate the initial kinetic energy.
      \[ \begin{aligned} KE_i &=\frac{1}{2}mv^2 \end{aligned} \]
    7. Calculate the final kinetic energy.
    8. Since the object comes to rest,
      \[v_f=0\]
    9. Therefore,
      \[ \begin{aligned} KE_f &=\frac{1}{2}m(0)^2\\ &=0 \end{aligned} \]
    10. Apply the Work-Energy Theorem.
      \[ \begin{aligned} W &=KE_f-KE_i\\ &=0-\frac{1}{2}mv^2\\ &=-\frac{1}{2}mv^2 \end{aligned} \]
    11. Therefore, the work done on the object is
      \[\boxed{W=-\frac{1}{2}mv^2}\]
    12. The negative sign indicates that the applied force acts opposite to the direction of motion, thereby removing the object's kinetic energy and bringing it to rest.
    13. If only the magnitude of the work required is asked, then
      \[\boxed{\left|W\right|=\frac{1}{2}mv^2}\]
    🔍 Physical Interpretation
    ›

    The object's initial kinetic energy is completely removed while bringing it to rest. This energy is usually transformed into other forms such as

    • Heat energy due to friction.
    • Sound energy.
    • Deformation energy in brakes or other stopping mechanisms.

    Thus, energy is conserved because the kinetic energy is transformed into other forms rather than being destroyed.

    🎯 Exam Significance
    Exam Significance
    ›
    • This is a direct application of the Work-Energy Theorem, one of the most important concepts in mechanics.
    • Students should remember that the work done while stopping an object is always negative because the object's kinetic energy decreases.
    • Questions involving the derivation of
      \[ W=-\frac{1}{2}mv^2 \]
      are frequently asked in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
    • This concept provides the foundation for understanding braking distance, collisions and conservation of energy in higher classes.
    • Competitive examinations often distinguish between the work done and the magnitude of work required.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  6 points

    1. Work done equals the change in kinetic energy.

    2. Initial kinetic energy is

      \[ \frac{1}{2}mv^2. \]

    3. Final kinetic energy is zero when the object comes to rest.

    4. The work done on the object is

      \[ -\frac{1}{2}mv^2. \]

    5. The negative sign indicates that the force opposes the motion.

    6. The kinetic energy of the object is transformed into other forms of energy such as heat and sound.

    ← Q15
    16 / 21  ·  76%
    Q17 →
    Q17
    NUMERIC3 marks
    Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h.
    📘 Concept & Theory
    Theory / Concept
    ›

    When a moving object is brought to rest, its kinetic energy becomes zero. According to the Work-Energy Theorem, the work done by the external force is equal to the change in the kinetic energy of the object.

    Mathematically,

    \[ W=\Delta KE \]

    Since kinetic energy is given by

    \[ KE=\frac{1}{2}mv^2 \]

    the work done in stopping the object is

    \[ W=KE_f-KE_i \]

    When the object comes to rest,

    \[ KE_f=0 \]

    Hence,

    \[ W=-\frac{1}{2}mu^2 \]

    The negative sign indicates that the stopping force acts opposite to the direction of motion. If only the amount (magnitude) of work required is asked, we calculate the loss of kinetic energy.

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Write the given data.

    2. Convert the speed from km/h to m/s.

    3. Calculate the initial kinetic energy.

    4. Apply the Work-Energy Theorem.

    5. State the answer with the correct SI unit.

    📊 Graph / Figure
    Graph / Figure
    ›
    Work Done by Braking Force Braking Distance (d) INITIAL STATE Moving Car u = 60 km/h Direction of Motion Braking Force (F) FINAL STATE Car Stops v = 0 km/h Work Done = Loss of Kinetic Energy (ΔKE) W = −2.08 × 105 J
    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  10 steps
    1. Given
    2. Mass of the car,
      \[m=1500\,\text{kg}\]
      Initial speed,
      \[u=60\,\text{km h}^{-1}\]
      Final speed,
      \[v=0\]
    3. Step-by-step Solution
    4. Convert the speed into SI units.
      \[ \begin{aligned} u &=60\times\frac{1000}{3600}\\ &=\frac{50}{3}\,\text{m s}^{-1}\\ &\approx16.67\,\text{m s}^{-1} \end{aligned} \]
    5. Calculate the initial kinetic energy.
      \[KE=\frac{1}{2}mu^2\]
    6. Substituting the values,
      \[ \begin{aligned} KE &=\frac{1}{2}\times1500\times\left(\frac{50}{3}\right)^2\\ &=750\times\frac{2500}{9}\\ &=\frac{1\,875\,000}{9}\\ &=208\,333.33\ \text{J} \end{aligned} \]
    7. Apply the Work-Energy Theorem.
    8. Since the car is brought to rest,
      \[KE_f=0\]
    9. Therefore,
      \[ \begin{aligned} W &=KE_f-KE_i\\ &=0-208\,333.33\\ &=-208\,333.33\ \text{J} \end{aligned} \]
    10. Thus, the work done by the braking force is
      \[\boxed{W=-2.08\times10^5\ \text{J}}\]
    11. If the question asks for the work required to stop the car (magnitude), then
      \[\boxed{\left|W\right|=2.08\times10^5\ \text{J}}\]
    12. The negative sign indicates that the braking force acts opposite to the direction of motion and removes the kinetic energy of the car.
    🎯 Exam Significance
    Exam Significance
    ›
    • This is a standard numerical problem based on the Work-Energy Theorem.
    • Students should always convert speed from km/h to m/s before substituting values into formulae.
    • The problem demonstrates that stopping a moving object requires removing its kinetic energy.
    • Similar numericals are frequently asked in CBSE Board examinations, Olympiads, NTSE and JEE Foundation-level examinations.
    • This concept is fundamental for understanding braking systems, road safety, momentum and collision problems in higher classes.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  5 points

    1. Convert velocity into SI units before calculations.

    2. Use

      \[ KE=\frac{1}{2}mv^2 \]
      to calculate the initial kinetic energy.

    3. The work done in stopping an object equals the decrease in its kinetic energy.

    4. The braking force performs negative work because it acts opposite to the direction of motion.

    5. The magnitude of work required to stop the car is

      \[ 2.08\times10^5\ \text{J}. \]
    ← Q16
    17 / 21  ·  81%
    Q18 →
    Q18
    NUMERIC3 marks
    In each of the following a force, \(F\) is acting on an object of mass \(m\). The direction of displacement is from west to east as shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.
    📘 Concept & Theory
    Theory / Concept
    ›

    The work done by a force depends not only on the magnitude of the force and the displacement but also on the angle between them.

    The mathematical expression for work done is

    \[ W=Fs\cos\theta \]

    where

    • \(W\) = Work done
    • \(F\) = Applied force
    • \(s\) = Displacement of the object
    • \(\theta\) = Angle between the force and the displacement

    Depending upon the value of \(\theta\), the work done can be

    • Positive when the force acts in the direction of displacement.
    • Negative when the force acts opposite to the displacement.
    • Zero when the force is perpendicular to the displacement.
    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Identify the direction of displacement.

    2. Determine the direction of the applied force.

    3. Find the angle between the force and displacement.

    4. Apply the work done formula.

    5. State whether the work done is positive, negative or zero.

    📊 Graph / Figure
    Graph / Figure
    ›
    Work Done in Different Situations Case 1: Zero Work m d F θ = 90° W = Fd cos(90°) = 0 Case 2: Positive Work m d F θ = 0° W = + Fd Case 3: Negative Work m d F θ = 180° W = − Fd
    Relationship between the direction of force and displacement determines the sign of work done.
    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  13 steps
    1. Case 1: Force Perpendicular to the Direction of Motion
    2. The force acts at right angles to the displacement.
    3. Therefore,
      \[\theta=90^\circ\]
    4. Using the work done formula,
      \[ \begin{aligned} W &=Fs\cos90^\circ\\ &=Fs\times0\\ &=0 \end{aligned} \]
    5. Work done is Zero.
    6. Case 2: Force in the Same Direction as the Motion
    7. The force and displacement are along the same direction.
    8. Hence,
      \[\theta=0^\circ\]
    9. Therefore,
      \[ \begin{aligned} W &=Fs\cos0^\circ\\ &=Fs\times1\\ &=Fs \end{aligned} \]
    10. Work done is Positive.
    11. Case 3: Force Opposite to the Direction of Motion
    12. The force acts opposite to the displacement.
    13. Therefore,
      \[\theta=180^\circ\]
    14. Hence,
      \[ \begin{aligned} W &=Fs\cos180^\circ\\ &=Fs\times(-1)\\ &=-Fs \end{aligned} \]
    15. Work done is Negative.
    16. Summary Table
    17. Case Angle Between Force and Displacement Work Done
      Force perpendicular to displacement \(90^\circ\) Zero
      Force in the direction of displacement \(0^\circ\) Positive
      Force opposite to displacement \(180^\circ\) Negative
    🎯 Exam Significance
    Exam Significance
    ›
    • This is one of the most important conceptual questions from the chapter Work and Energy.
    • Students should understand that the angle between force and displacement determines whether the work done is positive, negative or zero.
    • The concepts of positive, negative and zero work are frequently tested in CBSE Board examinations.
    • Similar conceptual questions regularly appear in Olympiads, NTSE, JEE Foundation and other competitive examinations.
    • These ideas form the basis for understanding the Work-Energy Theorem, conservative forces and circular motion in higher classes.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  5 points

    1. Work done depends on the angle between force and displacement.

    2. Force in the direction of motion produces positive work.

    3. Force opposite to the direction of motion produces negative work.

    4. Force perpendicular to the displacement does no work.

    5. Always use

      \[ W=Fs\cos\theta \]
      to determine the sign and magnitude of work done.

    ← Q17
    18 / 21  ·  86%
    Q19 →
    Q19
    NUMERIC3 marks
    Soni says that the acceleration of an object could be zero even when several forces are acting on it. Do you agree with her? Why?
    📘 Concept & Theory
    Theory / Concept
    ›

    According to Newton's Second Law of Motion, the acceleration produced in an object depends on the net (resultant) force acting on it, not on the number of forces acting on it.

    Mathematically,

    \[ F_{\text{net}}=ma \]

    where

    • \(F_{\text{net}}\) = Resultant (net) force acting on the object
    • \(m\) = Mass of the object
    • \(a\) = Acceleration produced

    If several forces act on an object but they exactly balance one another, then the resultant force becomes

    \[ F_{\text{net}}=0 \]

    Therefore,

    \[ \begin{aligned} F_{\text{net}} &=ma\ 0 &=ma \end{aligned} \]

    Since the mass of the object is non-zero,

    \[ \boxed{a=0} \]

    Thus, it is the resultant force, and not the number of forces, that determines the acceleration of an object.

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Recall Newton's Second Law of Motion.

    2. Determine whether the forces are balanced or unbalanced.

    3. Find the net force acting on the object.

    4. Relate the net force to acceleration.

    5. State the conclusion with justification.

    📊 Graph / Figure
    Graph / Figure
    ›
    Balanced Forces Produce Zero Acceleration Normal Force (F_N) Weight (mg) Friction (f) Applied Force (F_a) Object Net Force = 0 ⇒ Acceleration = 0
    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  9 steps
    1. Yes, I agree with Soni.
    2. An object can have zero acceleration even though several forces are acting on it, provided that all the forces are balanced.
    3. Step 1: Let several forces act on the object.
    4. If these forces cancel one another, then the resultant force becomes
      \[F_{\text{net}}=0\]
    5. According to Newton's Second Law,
      \[F_{\text{net}}=ma\]
    6. Substituting \(F_{\text{net}}=0\),
      \[ \begin{aligned} 0 &=ma\\ a &=0 \end{aligned} \]
    7. Therefore, the acceleration of the object is zero.
    8. Under this condition, the object
      • Remains at rest if it was initially at rest.
      • Continues to move with constant velocity if it was already in motion.
    9. Hence, zero acceleration is possible even when several forces act on an object, provided their resultant is zero.
    🎯 Exam Significance
    Exam Significance
    ›
    • This question tests the understanding of balanced and unbalanced forces.
    • Students should remember that acceleration depends on the net force, not on the number of forces acting.
    • Conceptual questions based on Newton's Second Law are frequently asked in CBSE Board examinations.
    • Similar reasoning-based questions appear in Olympiads, NTSE, JEE Foundation and other competitive examinations.
    • This concept provides the basis for studying equilibrium, mechanics and dynamics in higher classes.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  5 points

    1. Acceleration depends on the resultant (net) force.

    2. Balanced forces produce zero net force.

    3. If

      \[ F_{\text{net}}=0, \]
      then
      \[ a=0. \]

    4. An object with zero acceleration may either remain at rest or move with constant velocity.

    5. The number of forces acting on an object does not determine its acceleration; only their resultant does.

    ← Q18
    19 / 21  ·  90%
    Q20 →
    Q20
    NUMERIC3 marks
    Find the energy in joules consumed in 10 hours by four devices of power 500 W each.
    📘 Concept & Theory
    Theory / Concept
    ›

    Power is the rate at which electrical energy is consumed or work is done.

    It is defined as

    \[ P=\frac{E}{t} \]

    Therefore,

    \[ E=P\times t \]

    where

    • \(P\) = Power (W)
    • \(E\) = Energy consumed (J)
    • \(t\) = Time (s)

    When several electrical devices operate simultaneously, their powers are added to obtain the total power.

    Since power is measured in watts, time should be converted into seconds to obtain the energy in joules.

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Calculate the total power of all four devices.

    2. Convert the operating time into seconds.

    3. Apply the relation \(E=P\times t\).

    4. Calculate the energy consumed.

    5. Express the answer in scientific notation.

    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  9 steps
    1. Given
    2. Power of each device,
      \[ 500\ \text{W}\]
    3. Number of devices,
      \[4\]
    4. Time of operation,
      \[10\ \text{hours}\]
    5. Step-by-step Solution
    6. Calculate the total power.
      \[ \begin{aligned} P &=4\times500\\ &=2000\ \text{W} \end{aligned} \]
    7. Convert the time into seconds.
      \[ \begin{aligned} t &=10\times60\times60\\ &=36000\ \text{s} \end{aligned} \]
    8. Apply the formula
      \[ E=P\times t \]
    9. Substituting the values,
      \[ \begin{aligned} E &=2000\times36000\\ &=72\,000\,000\ \text{J} \end{aligned} \]
    10. Express the answer in scientific notation.
      \[ \begin{aligned} E &=72\,000\,000\ \text{J}\\ &=7.2\times10^7\ \text{J} \end{aligned} \]
    11. Therefore, the total electrical energy consumed by the four devices in 10 hours is
      \[ \boxed{7.2\times10^7\ \text{J}} \]
    🎯 Exam Significance
    Exam Significance
    ›
    • This question tests the relationship between power, time and electrical energy.
    • Students should remember to add the powers of all devices before calculating the total energy consumed.
    • Always convert time into seconds when the answer is required in joules.
    • Similar numerical problems are frequently asked in CBSE Board examinations, NTSE, Olympiads and JEE Foundation-level examinations.
    • This concept is useful in understanding electricity bills, household power consumption and energy conservation.
    ← Q19
    20 / 21  ·  95%
    Q21 →
    Q21
    NUMERIC3 marks
    A freely falling object eventually stops on reaching the ground. What happens to its kinetic energy?
    📘 Concept & Theory
    Theory / Concept
    ›

    According to the Law of Conservation of Energy, energy can neither be created nor destroyed. It can only be transformed from one form into another.

    During the free fall of an object, its gravitational potential energy is continuously converted into kinetic energy.

    Just before the object strikes the ground,

    • Potential energy is minimum.
    • Kinetic energy is maximum.

    When the object collides with the ground, its velocity becomes zero. Hence, its kinetic energy also becomes zero. However, this energy does not disappear; it is transformed into other forms of energy.

    🗺️ Solution Roadmap
    Step-by-step Plan
    ›

    1. Describe the energy of the object just before striking the ground.

    2. Explain what happens during the collision.

    3. Identify the different forms into which the kinetic energy is transformed.

    4. Relate the process to the Law of Conservation of Energy.

    5. State the conclusion.

    📊 Graph / Figure
    Graph / Figure
    ›
    Fig. 1 — Free body diagram
    Fig. 1 — Free body diagram
    ✏️ Solution
    Complete Solution
    ›
    Step-by-step Solution  ·  8 steps
    1. As a freely falling object approaches the ground, its gravitational potential energy is continuously converted into kinetic energy. Consequently, just before it strikes the ground, its kinetic energy is maximum.
    2. During the impact with the ground, the object comes to rest.
    3. Therefore,
      \[v=0\]
    4. Hence,
      \[ \begin{aligned} KE &=\frac{1}{2}mv^2\\ &=\frac{1}{2}m(0)^2\\ &=0 \end{aligned} \]
    5. The kinetic energy does not disappear. Instead, it is transformed into other forms of energy such as
      • Heat energy produced due to the impact and friction between the object and the ground.
      • Sound energy produced during the collision.
      • Elastic potential energy if the object or the ground undergoes temporary deformation.
      • Deformation energy if the object or the ground is permanently deformed.
    6. Therefore, although the object stops moving, its kinetic energy is converted into heat, sound and deformation energy.
    7. Thus, the Law of Conservation of Energy is fully satisfied because the total energy remains constant; only its form changes.
    8. Energy Transformation
    9. \[ \boxed{ \text{Potential Energy} \rightarrow \text{Kinetic Energy} \rightarrow \text{Heat Energy} + \text{Sound Energy} + \text{Deformation Energy} } \]
    🎯 Exam Significance
    Exam Significance
    ›
    • This is a frequently asked conceptual question based on the Law of Conservation of Energy.
    • Students should remember that energy is never destroyed when an object stops; it is transformed into other forms.
    • Questions involving energy transformation during collisions are common in CBSE Board examinations.
    • Similar conceptual questions are asked in Olympiads, NTSE, JEE Foundation and other competitive examinations.
    • This concept forms the foundation for studying collisions, impact mechanics and energy conservation in higher classes.
    🔑 Key Takeaways
    Key Takeaways
    ›
    Key Takeaways  ·  5 points

    1. During free fall, potential energy is converted into kinetic energy.

    2. On striking the ground, the object's kinetic energy becomes zero because its velocity becomes zero.

    3. The kinetic energy is transformed into heat, sound and deformation energy.

    4. No energy is lost; only its form changes.

    5. The process is a practical demonstration of the Law of Conservation of Energy.

    ← Q20
    21 / 21  ·  100%
    ↑ Back to top
    🎓

    Chapter Complete!

    All 21 solutions for Work and Energy covered.

    ↑ Review from the top
    📚
    ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
    Sharing this chapter
    NCERT Class 9 Work and Energy Exercise Solutions
    NCERT Class 9 Work and Energy Exercise Solutions — Complete Notes & Solutions · academia-aeternum.com
    Work and energy are two fundamental concepts in physics that explain how things move and change in the world around us. This chapter of NCERT Class 9 Science explores the relationship between force, motion, and energy through engaging examples and practical problems. Students learn how work is done when a force causes displacement, the different forms of energy such as kinetic and potential energy, and the principle of conservation of energy. Our detailed NCERT Class 9 Work and Energy textbook…
    🎓 Class 9 📐 Science 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
    Share on
    WhatsApp
    Telegram
    X / Twitter
    Facebook
    Instagram
    Post to Instagram
    Threads
    Pinterest
    LinkedIn
    Quora
    Discord
    Reddit
    Email
    academia-aeternum.com/class-9/science/work-and-energy/exercises/ Copy link
    💡
    Exam tip: Sharing chapter notes with your study group creates a reinforcement loop. Teaching a concept is the fastest path to mastering it.

    Recent posts

      Force and Laws of Motion — Learning Resources

      📄 Detailed Notes
      🧠 Practice MCQs
      ✔️ True / False

      Frequently Asked Questions

      Work is energy transfer due to displacement by force; energy is the ability to perform work.

      It stores chemical energy and converts it to electrical energy.

      Energy flows from the sun to producers, then to consumers via food chain.

      The rate at which the appliance consumes energy per unit time.

      Lighting a bulb.

      Potential energy converts to kinetic energy.

      It guides efficient energy usage and device design.

      No, according to the law of conservation of energy.

      Light energy from the sun.

      Get in Touch

      Let's Connect

      Questions, feedback, or suggestions?
      We'd love to hear from you.