WORK AND ENERGY-Notes

work is said to be done when a force is applied on an object and that object moves in the direction of the applied force. If there is no movement, or if the movement is not in the direction of the force, then no work is said to be done.

In simpler words, work happens only when both force and displacement act together in the same direction. For example, if you push a box and it actually moves, you are doing work. But if you push hard and the box doesn’t move, no work is done in the scientific sense — even though you may feel tired.

Work can be positive or negative depending on the direction of force and displacement.

Work = Force × Displacement in the direction of force
Thus, work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. Work has only magnitude and no direction.

if \(F = 1\, N\) and \(s = 1\, m\) then the work done by the force will be \(1\, N\, m\). Here the unit of work is newton metre \((N\, m)\) or joule (J).

Thus, 1 J is the amount of work done on an object when a force of 1 N displaces it by 1 m along the line of action of the force.

A force of 7 N acts on an object. The displacement is, say 8 m, in the direction of the force Let us take it that the force acts on the object through the displacement. What is the work done in this case?

Solution:
Force \(F=7\,N\)
Displacement in the direction of Force \(d=8\,m\)
Work done \(W=\) \[ W=F\cdot d\tag{1}\\ \] SUbstituting Values of \(F\) and \(d\) in Formula {1} \[ \begin{align} W=7\times 8\\=56\,Nm \end{align} \]

A porter lifts a luggage of 15 kg from the ground and puts it on his head 1.5 m above the ground. Calculate the work done by him on the luggage.

Solution:
Weight of the luggage \((m)=15\,kg\)
Force acting on the luggage due to gravity \[\begin{aligned}F&=m\times g\\&=15\times 10\\&=150\,N\end{aligned}\] Displacement in the direction of Force \((d)=1.5\,m\)
Work done \(W=\) \[ W=F\cdot d\tag{1}\\ \] Substituting Values of \(F\) and \(d\) in Formula {1} \[ \begin{align} W&=150\times 1.5\\&=225\,Nm \end{align} \]

Energy is defined as the capacity of a body to do work.

The unit of energy is, therefore, the same as that of work, that is, joule (J). 1 J is the energy required to do 1 joule of work.
Sometimes a larger unit of energy called kilo joule (kJ) is used. 1 kJ equals 1000 J.

Energy exists in several forms, each characterized by how it manifests and operates in nature and technology

Major Forms of Energy

• Kinetic Energy:
  The energy possessed by an object due to its motion. Every moving object, like a rolling ball or a running person, possesses kinetic energy because of movement.
• Potential Energy:
  Energy stored in an object due to its position or shape, such as a stretched rubber band or water held behind a dam. Varieties include gravitational, elastic, and chemical potential energy.
• Mechanical Energy:
  The sum of kinetic and potential energy present in a body, representing the total energy from motion and position.
• Chemical Energy:
  The energy stored in chemical bonds of substances, released during chemical reactions—examples include energy in batteries or food.
• Electrical Energy:
  The energy resulting from the movement of electric charges, such as the power flowing through electric wires to light a bulb.
• Heat (Thermal) Energy:
  Energy related to an object's temperature, produced through the movement of its particles. A hot cup of tea demonstrates heat energy.
• Light (Radiant) Energy:
  Energy carried by light waves or electromagnetic radiation—essential for vision and many forms of technology.
• Sound Energy:
  The energy produced by vibrating objects, responsible for what we hear as sound.
• Nuclear Energy:
  The energy released from atomic nuclei during processes like fission or fusion, powering nuclear reactors and stars.

Kinetic energy is the energy that a body possesses due to its motion. Any object that is moving, whether it's a rolling ball, a flying airplane, or a running person, has kinetic energy. The amount of kinetic energy an object has depends on its mass and the square of its velocity.

Mathematically, kinetic energy (KE) is expressed as: \[KE=\frac{1}{2}mv^2\] Where,
\(m\) is the mass of the object
\(v\) is the velocity of the object

Derivation

From Equation of motion, we know that \[\begin{align}v^2-u^2&=2as\\ \implies s&=\frac{v^2-u^2}{2a}\tag{1}\end{align}\] Force \[F=ma\tag{2}\] Work done \[ W=Fs\tag{3} \] Substituting values for \(s\) and \(F\) from equation (1) and (2) into (3) \[ \require{cancel} \begin{aligned} W&=ma\left(\frac{v^2-u^2}{2a}\right)\\\\ &=m\cancel{a}\left(\frac{v^2-u^2}{2\cancel{a}}\right)\\\\ &=\frac{1}{2}m({v^2-u^2}) \end{aligned} \] If the object is starting from its stationary position, that is, \(u = 0\), the \[W=\frac{1}{2}mv^2\] Thus, the kinetic energy possessed by an object of mass, \(m\) and moving with a uniform velocity, \(v\) is \[\boxed{E_k=\frac{1}{2}mv^2}\]

An object of mass 15 kg is moving with a uniform velocity of \(4\, m s^{–1}\). What is the kinetic energy possessed by the object?

Solution:
Mass of the object=15 kg
Uniform Velocity = \(4\, m s^{–1}\)
Kinetic Energy \(E_k=\) \[E_k=\frac{1}{2}mv^2\] Substituting Values of \(m\) and \(v\) \[\small\begin{aligned}E_k&=\frac{1}{2}\times 15\,kg\times4\,ms^{-1}\times4\,ms^{-1}\\\\ &=120\,\mathrm{J}\end{aligned}\]

What is the work to be done to increase the velocity of a car from \(30\, km h^{–1}\) to \(60\, km h^{–1}\) if the mass of the car is 1500 kg?

Solution:
Initial Velocity of the car \[\begin{aligned}u=30\,kmh^{-1}\\\\ =\frac{(30\times 1000) \,m}{(60\times 60)\;s}\\\\ =\frac{50}{3}ms^{-1} \end{aligned}\] Final Velocity car \[\begin{aligned}v=60\,kmh^{-1}\\\\ =\frac{(60\times 1000) \,m}{(60\times 60)\;s}\\\\ =\frac{25}{3}ms^{-1} \end{aligned}\] Mass of the car \(m=1500\,kg\) Work done to increase velocity \[ \scriptsize \require{cancel} \begin{aligned} W&=\frac{1}{2}m(v^2-u^2)\\\\ &=\frac{1}{2}\times 1500\left[\left(\frac{50}{3}\right)^2-\left(\frac{25}{3}\right)^2\right]\\\\ &=\frac{1}{2}\times 1500\left[\left(\frac{50}{3}+\frac{25}{3}\right)\left(\frac{50}{3}-\frac{25}{3}\right)\right]\\\\ &=\frac{1}{2}\times 1500\left[\left(\frac{75}{3}\right)\left(\frac{25}{3}\right)\right]\\\\ &=\frac{1}{2}\times \cancelto{5}{15}00\times \frac{\cancelto{25}{75}}{\cancel{3}} \times \frac{25}{\cancel{3}}\\\\ &=\frac{1}{\cancel{2}}\times\cancelto{250}{500}\times 625\\\\ &=156250\,\mathrm{J} \end{aligned}\]

Potential energy is the energy stored in an object due to its position or configuration. It represents the capacity to do work that an object possesses when it is in a particular state or position, rather than from its motion.

For example, when a brick is lifted to a certain height from the ground, energy is spent in raising it, and this energy remains stored as potential energy because of the brick's elevated position. If the brick is released, this stored energy can convert to other forms, like kinetic energy.

There are mainly two common types of potential energy studied in Class IX

• Gravitational Potential Energy:
  Energy possessed by an object due to its height above the ground. The higher the object, the greater the gravitational potential energy, calculated as \[PE=m\times g\times h\] where \(m\) is mass, \(g\) is acceleration due to gravity, and \(h\) is height.
• Elastic Potential Energy:
  Energy stored in objects that can be stretched or compressed, like a stretched bow or a compressed spring.

Potential energy depends on position or shape, and it can transform into kinetic energy when the object moves. Its SI unit is the joule (J), same as for all forms of energy. This concept teaches how energy can be stored and later released to perform work.

Find the energy possessed by an object of mass 10 kg when it is at a height of \(6\, m\) above the ground. Given, \(g = 9.8 m s^{–2}\).

Solution:
mass of the object \(=10\, k\)g
Height of the object where it is placed \(= 6\,m\)
Acceleration due to gravity \(=9.8\,m/s^2\)
Calculation of Potential Energy \(PE:\) $$\begin{aligned}PE&=mgh\\ &=10\times 9.8\times 6\\ &=10\times 58.8\\ &=588\,\mathrm{J}\end{aligned}$$

An object of mass 12 kg is at a certain height above the ground. If the potential energy of the object is 480 J, find the height at which the object is with respect to the ground. Given, \(g = 10\,m s^{–2}\).

Solution:
mass of the object = 12kg
PE = 480J
Acceleration due to gravity \(= 10\,ms^{-2}\)
Let height at which objects placed be \(h\)
$$\begin{aligned}PE&=mgh\\ 480&=12\times 10\times h\\\\ \Rightarrow h&=\dfrac{480}{12\times 10}\\\\ &=4m\end{aligned}$$

The law of conservation of energy states that energy can neither be created nor destroyed; it can only be transformed from one form to another.

This means that the total energy in an isolated system remains constant over time.

Energy may change its form — for example, from potential energy to kinetic energy or from chemical energy to thermal energy — but the overall amount of energy does not increase or decrease.

potential energy + kinetic energy = constant \[PE+KE=\text{ constant}\\\\\text{or}\\ \boxed{mgh+\frac{1}{2}mv^2=\text{ constant}}\]

The sum of kinetic energy and potential energy of an object is its total mechanical energy.

during the free fall of the object, the decrease in potential energy, at any point in its path, appears as an equal amount of increase in kinetic energy.

Illustration

An object of mass 20 kg is dropped from a height of 4 m. Fill in the blanks in the following table by computing the potential energy and kinetic energy in each case.

Solution:
mass of object = 20kg
Height at which object is placed = 4m
Initial Velocity \(u=0\)

$$\begin{aligned}PE_{\left( 4\right) }&=mgh\\ &=20\times 10\times 4\\ &=800J\\\\ v&=0\\ KE_{\left( 4\right) }&=\dfrac{1}{2}mv^{2}\\ &=\dfrac{1}{2}m\left( 0\right) ^{2}\\ &=0\\\\ PE_{\left( 3\right) }&=2\times 10\times 3\\ &=600J\\\\ v^{2}-u^{2}&=2as\\ v^{2}-0&=2\cdot 10\cdot 1\\ v^{2}&=20\\ KE_{(3)}&=\frac{1}{2}mv^{2}\\ &=\dfrac{1}{2}\times 20\times 20\\ &=200J\\\\ PE_{\left( 2\right)} &=2\times 10\times 2\\ &=400J\\\\ v^{2}-u^{2}&=2\times 10\times 2\\ v^{2}&=40\\ KE_{(2)}&=\frac{1}{2}mv^{2}\\ &=\frac{1}{2}\times 20\times 40\\ &=400J\\\\ PE_{\left( 1\right)} &=20\times 10\times 1\\ &=200J\\\\ v^{2}&=2\times 10\times 3\\ &=60\\ KE_{(1)}&=\dfrac{1}{2}\times 20\times 60\\ &=600J\\\\ PE_{\left( 0\right) }&=0\\ v^{2}&=2\times 10\times 4\\ &=80\\ KE_{(0)}&=\dfrac{1}{2}\times 20\times 80\\ &=800J\end{aligned}$$

The rate of doing work is called power in physics. It measures how quickly work is done or energy is transferred over time. Mathematically, power \((P)\) is defined as the amount of work done \((W)\) divided by the time \((t)\) taken to do that work: \[\boxed{P=\dfrac{W}{t}}\] The SI unit of power is the watt \((W)\), [in honour of James Watt (1736 – 1819)] where one watt equals one joule per second \((1\, W = 1\, J/s)\).

Two girls, each of weight 400 N, climb up a rope to a height of 8 m. We name one of the girls A and the other B. Girl A takes 20 s while B takes 50 s to accomplish this task. What is the power expended by each girl

Solution:
Weight of girls \(mg= 400\,N\)
Height to climb \(h= 8\,m\)
Time taken by Girl A \(t= 20\,s\)
Time taken by Girl B \(t= 50\,s\)
To find Power Extended by Girls

Work done $$\begin{aligned}W&=mgh\\ W&=400\times 8\\ &=3200J\end{aligned}$$ Power Extended by Girl A $$\begin{aligned}P&=\dfrac{W}{t}\\\\ &=\dfrac{3200}{20}\\\\ &=160\,\mathrm{W}\end{aligned}$$ Power Extended by Girl B $$\begin{aligned}P&=\dfrac{W}{t}\\\\ &=\dfrac{3200}{50}\\\\ &=64\,\mathrm{W}\end{aligned}$$

A boy of mass 50 kg runs up a staircase of 45 steps in 9 s. If the height of each step is 15 cm, find his power. Take \(g = 10 \,m s^{–2}\).

Solution:
Mass of the boy \((m)= 50\,kg\)
Acceleration due to gravity \((g)= 10\,ms^{-2}\)
Height = 45×15 cm = 675cm \((h)= 6.75\,m\)
Time taken \((t)= 9\,s\)
$$\begin{aligned}W&=mgh\\ &=50\times 10\times 67.5\\ &=3375\,J\\\\ P&=\dfrac{W}{t}\\\\ &=\dfrac{3375}{9}\\\\ &=375\,W\end{aligned}$$

• Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force. The unit of work is joule: 1 joule = 1 newton × 1 metre.
• Work done on an object by a force would be zero if the displacement of the object is zero.
• An object having capability to do work is said to possess energy. Energy has the same unit as that of work.
• An object in motion possesses what is known as the kinetic energy of the object. An object of mass, m moving with \(\frac{1}{2}mv^2\).
• The energy possessed by a body due to its change in position or shape is called the potential energy. The gravitational potential energy of an object of mass, m, raised through a height, h, from the earth’s surface is given by \(m g h\).
• According to the law of conservation of energy, energy can only be transformed from one form to another; it can neither be created nor destroyed. The total energy before and after the transformation always remains constant.
• Energy exists in nature in several forms, such as kinetic energy, potential energy, heat energy, chemical energy, etc. The sum of the kinetic and potential energies of an object is called its mechanical energy.
• Power is defined as the rate of doing work. The SI unit of power is the watt. \(1\, W = 1 \,J/s\).