WORK AND ENERGY-Exercise

Q1. 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 wind-mill 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

Solution:

• Suma is swimming in a pond:
  Work is done because Suma applies force to push water backward to move forward, causing displacement in the direction of the force.
• A donkey is carrying a load on its back:
  No work is done by the donkey on the load in the physical sense because the load does not move in the direction of the applied force (the load is stationary relative to the donkey).
• A wind-mill is lifting water from a well:
  Work is done because the wind-mill applies force to lift water, causing displacement of water in the direction of the force.
• A green plant is carrying out photosynthesis:
  No work is done in the physics sense because there is no mechanical force causing displacement.
• An engine is pulling a train:
  Work is done since the engine exerts force that causes the train to move in the direction of the force.
• Food grains are getting dried in the sun:
  No work is done mechanically because there is no force causing displacement of the grains.
• A sailboat is moving due to wind energy:
  Work is done as the wind exerts force on the sail causing the boat to move, resulting in displacement in the direction of the force.

Q2. 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?

Solution:
When an object is thrown at an angle and moves along a curved path in projectile motion, it comes back to the ground on the same horizontal level from where it was projected. The work done by the force of gravity on such an object over the entire motion is zero.

This is because work done by a force is given by: \[W=Fd\] Where \(F\) is force and \(d\) is displacement.

In projectile motion under gravity:

• In projectile motion under gravity
• The vertical displacement of the object after completing the motion is zero since it returns to the same horizontal level

Q3. A battery lights a bulb. Describe the energy changes involved in the process.

Solution:
When a battery lights a bulb, a series of energy transformations take place to produce light and heat.

Initially, the battery contains stored chemical energy within its internal chemicals. When connected in a circuit, these chemical reactions convert chemical energy into electrical energy. This electrical energy flows through the circuit wires to the bulb.

Inside the bulb, electrical energy flows through the filament, which resists the flow of electricity, causing the filament to heat up. This heating converts electrical energy into heat energy. As the filament becomes very hot, it emits light energy, making the bulb glow.

So, the energy changes involved can be described as:

Chemical energy (in battery) \(\Rightarrow\) Electrical energy (in circuit) \(\Rightarrow\) Heat energy + Light energy (in bulb)

This process is a classic example of energy conversion, where one form of energy transforms into others while obeying the law of conservation of energy. The heat generated also explains why bulbs get warm after being lit for some time.

Thus, the battery’s chemical energy ultimately becomes useful light energy, along with some heat energy, through electrical energy as an intermediate step.

Q4. 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.

Solution:
Mass \(20\) kg
Initial Velocity \(u=5\,m/s\)
Final Velocity \(v=2\,m/s\)
Work Done \(W\): \[ \begin{aligned} W&=\frac{1}{2}m(v^2-u^2)\\\\ &=\frac{1}{2}\times20\times(2^2-5^2)\\\\ &=\frac{1}{2}\times20\times(4-25)\\\\ &=10\times(-21)\\\\ &=-210\,\mathrm{J} \end{aligned} \]

Q5. 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.

Solution:
When a 10 kg mass is moved from point A to point B on a table, and the line joining these points is horizontal, the work done by the gravitational force on the mass is zero. This is because gravitational force acts vertically downward, while the displacement of the object is purely horizontal.

Work done by a force depends on the component of displacement in the direction of the force. Mathematically, work is: \[W=F×d×cos\,\theta\] where
\(F\) is the force,
\(d\) is the displacement, and
\(\theta\) is the angle between the force and displacement. Since the force of gravity acts vertically downward and the displacement is horizontal, the angle \(\theta=90^\circ\), As \(cos\,90^\circ=0\) \[W=F\times d\times 0=0\] Therefore, gravity does not do any work when an object moves horizontally because it does not cause any displacement in its own direction. The energy of the object due to gravity remains unchanged in such horizontal movements.

This explanation highlights that work requires displacement along the force's direction, which is absent in purely horizontal movement against gravity. Hence, the work done by the gravitational force in this scenario is zero.

Q6. The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

Solution:
The potential energy of a freely falling object decreases progressively as it moves closer to the ground. However, this does not violate the law of conservation of energy. The key reason is that while the potential energy is reducing, the kinetic energy of the object increases by exactly the same amount.

Q7. What are the various energy transformations that occur when you are riding a bicycle?

Solution:
When you ride a bicycle, several energy transformations take place. Initially, the chemical energy stored in your body from food is converted into muscular energy. This muscular energy enables your legs to pedal the bicycle. As you pedal, your muscular energy changes into mechanical energy, which moves the bicycle parts, and finally into kinetic energy, which makes the bicycle and you move forward. Additionally, some energy is lost as heat due to friction between the bicycle's moving parts and air resistance, warming your body a little. So, the chain of energy changes is:

Chemical energy (food) \(\Rightarrow\) Muscular energy \(\Rightarrow\) Mechanical energy \(\Rightarrow\) Kinetic energy + Heat energy

This smooth flow of energy keeps you cycling efficiently while some energy dissipates as heat during the ride.

Q8. 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?

Solution:
When we push a huge rock with all our might but fail to move it, no work is done on the rock because work requires displacement in the direction of the force applied. Since the rock does not move, the energy we put in does not transfer to the rock as mechanical work.

However, the energy we spend is converted into other forms, mostly heat energy in our muscles due to muscular effort and friction within our body. This is why we feel tired and our body warms up even though the rock hasn't moved. So, while no energy is transferred to the rock as work, our body still uses energy and transforms it internally to sustain the effort.

Q9. A certain household has consumed 250 units of energy during a month. How much energy is this in joules?

Solution:
Total Units consumed during the month =250 1 unit of Electricity =1kWh, therefore \[\small\begin{aligned}250\,\text{units} &= 250\times 1\,kWh\\ &=250\times 1000\times 60\times 60\\ &=9\times 10^8\,\mathrm{J} \end{aligned}\]

Q10. 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.

Solution:
Mass of the Object = 40 kg Height to that it s raised = 5m \[ \begin{aligned} PE&=mgh\\ &=40\times 10\times 5\\ &=2000\,\mathrm{J} \end{aligned} \] Object is allowed to fall, KE at the height of 2.5m
Let the velocity of the object at 2.5 m is \(v\,m/s\)
Initial Velocity \(u=0\) \[ \begin{aligned} v^2&=u^2+2gs\\ v^2&=0+2\times 10\times 2.5\\ v^2&=50\\\\ KE&=\frac{1}{2}mv^2\\\\ &=\frac{1}{2}40\times 50\\\\ &=1000\,\mathrm{J} \end{aligned} \]

Q11. What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Solution:

Q12. Can there be displacement of an object in the absence of any force acting on it? Think. Discuss this question with your friends and teacher.

Solution:
Yes, an object can have displacement even if no force acts on it. According to Newton’s first law, an object in motion continues to move with constant velocity unless acted upon by an external force. So, if an object is already moving, it will keep moving and have displacement despite no force acting on it.

However, if the object starts from rest, it needs a force to set it in motion. Once moving, no additional force is required for displacement if friction and other resistances are absent or negligible.

Thus, displacement can happen without force during uniform motion, but force is needed to change the state of motion or to start moving.

Q13. 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.

Solution:
When a person holds a bundle of hay over their head for 30 minutes and gets tired, physically, they have not done any work on the hay from a physics standpoint. This is because work requires displacement in the direction of the applied force. Since the hay is stationary and does not move, the displacement is zero, so the mechanical work done on the hay is zero.

However, the person still expends energy internally because their muscles undergo metabolic activity to maintain that position against gravity. This energy is converted into heat and causes fatigue, but it is not considered mechanical work on the hay.

In summary, no mechanical work is done on the stationary hay since there is no displacement, but the person’s body uses energy to hold it up, leading to tiredness.

Q14. An electric heater is rated 1500 W. How much energy does it use in 10 hours?

Solution:
Heater is rated as 1500W
Energy consumption in 10 Hrs \[ \begin{aligned} P&=\frac{W}{t}\\\\ W&=P\times t\\\\ &=1500\times 10\times 60\times 60\\ &=54000000\\&=5.4\times10^7 \mathrm{J} \end{aligned} \]

Q15. 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?

Solution:
When we pull a pendulum bob to one side and release it, the bob starts oscillating, continuously converting energy between two forms. At its highest points, the bob has maximum potential energy and minimal kinetic energy because it is momentarily at rest. As it swings downward, potential energy transforms into kinetic energy, reaching a maximum at the lowest point of the swing, where speed is greatest and potential energy is minimum. As it rises on the other side, kinetic energy again changes back into potential energy.

The law of conservation of energy is illustrated here because the total mechanical energy — the sum of kinetic and potential energies — remains constant throughout the motion, assuming no external forces like air resistance.

However, the pendulum bob eventually comes to rest due to air resistance and friction at the pivot, which convert mechanical energy into heat and sound, spreading energy into the surroundings. This is not a violation of the conservation law; instead, the energy is simply transformed into other less useful forms, dissipating from the pendulum system.

Thus, the pendulum exemplifies energy conservation in an ideal system, while real-world factors cause energy loss and bring the oscillations to a stop.

Q16. 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?

Solution:
To bring an object of mass \(m\) moving at a constant velocity \(v\) to rest, the work done on the object should exactly remove its kinetic energy.

The kinetic energy (KE) of the object is: \[KE=\frac{1}{2}mv^2\] To stop the object, work done \(W\) must be equal in magnitude but opposite in sign to this kinetic energy. Hence: \[W=-\frac{1}{2}mv^2\] The negative sign indicates that work is done against the motion to bring the object to rest. Thus, the amount of work required to stop the object is \(\frac{1}{2}mv^2\), done in such a way as to remove its kinetic energy.

Q17. Calculate the work required to be done to stop a car of 1500 kg moving at a velocity of 60 km/h?

Solution:
Mass of the car \((m)=1500\,\mathrm{kg}\)
Initial Velocity \((u)=60\,\mathrm{km/h} =\frac{60\times 1000}{60\times60}\,\mathrm{m/s}\)
Final Velocity \((v)=0\)
Work Done to stop the car is equal to the Kinetic Energy of the car \[ \scriptsize \begin{aligned} KE&=\frac{1}{2}mv^2\\\\ &=\frac{1}{2}\times 1500\times\left(\frac{60\times 1000}{60\times60}\,\mathrm{m/s}\right)^2\\\\ &=750\times\frac{50}{3}\times\frac{50}{3}\\\\ &=\frac{250\times50\times50 }{3}\\\\ &=208333.33\,\mathrm{J} \end{aligned} \]

Q18. 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 shown by the longer arrow. Observe the diagrams carefully and state whether the work done by the force is negative, positive or zero.

Solution:
Case -1: When Force is perpendicular to the direction of Motion:
Work Done is Zero, As - Angle between Force and Motion is \(90^\circ\) \[ \begin{aligned} W&=F\,d\,cos\,90^\circ\\ &=F\times d\times 0\\ &=0 \end{aligned} \] Case -2: When Force and direction of motion is same:
Work Done is Positive, As - Angle between Force and Motion is \(0^\circ\) \[ \begin{aligned} W&=F\,d\,cos\,0^\circ\\ &=F\times d\times 1\\ &=F\,d \end{aligned} \] Case -3: When Force and direction of motion is opposite:
Work Done is Negative, As - Angle between Force and Motion is \(180^\circ\) \[ \begin{aligned} W&=F\,d\,cos\,180^\circ\\ &=F\times d\times (-1)\\ &=-F\,d \end{aligned} \]

Q19. Soni says that the acceleration in an object could be zero even when several forces are acting on it. Do you agree with her? Why?

Solution:
Yes, I agree with Soni. The acceleration of an object can be zero even when several forces are acting on it, as long as these forces balance each other out, resulting in a net force of zero.

According to Newton’s second law \[F_{net}=ma\] if the net force \(F_{net}\) is zero, then acceleration \(a\) must also be zero. This means the object either remains at rest or continues to move with constant velocity despite multiple forces acting on it, provided they cancel each other completely.

Hence, zero acceleration with multiple forces is possible when forces are balanced.

Q20. Find the energy in joules consumed in 10 hours by four devices of power 500 W each.

Solution:
Energy Consumption= 500 W/Device
Power of four devices \(=500\times 4=2000W\)
Energy consumed in 10 Hr = \[ E=P\times t\\ E=2000\,\mathrm{Watt}\times 10\,\mathrm{h}\times 60\,\mathrm{s}\times 60\,\mathrm{s}\\ =72000000\\ =7.2\times 10^7\,\mathrm{J} \]

Q21. A freely falling object eventually stops on reaching the ground. What happenes to its kinetic energy?

Solution:
When a freely falling object reaches the ground and stops, its kinetic energy does not vanish; instead, it transforms into other energy forms.

Most of this kinetic energy converts into heat energy due to the impact and friction at the point of contact, causing a slight rise in temperature. Simultaneously, some energy converts into sound energy produced by the impact noise.

Additionally, depending on the surface and the object, some energy may deform the object or ground, storing energy as potential energy in the deformation.

Thus, the kinetic energy spreads into heat, sound, and mechanical deformation energy, ensuring the total energy is conserved and no energy is lost from the system.