 # Derivation of Work Energy Theorem

## Work Energy Theorem

In our day-to-day life, work is defined as the act to do something. For example, if you are reading this article you are doing some work according to you. Physics, on the other side, doesn't agree with this statement because you remain still in your position, and there is no displacement of body in physical sense. Thus, the work-energy theorem describes the reasons behind this Physics of no work.

Work is said to be done when force acting on an object, displaces the object. If no displacement of the object occurs, work is not done. You may feel tired by standing for a long time, but according to Physics you have done no work.

Thus, work is the product of force and displacement. We know that moving objects possess kinetic energy. This establishes a relation between work and kinetic energy, which is called Work-Energy Theorem.” It is expressed as:

W = ΔK

Where,

W = work done in joules (J), and

ΔK = change in kinetic energy of the object.

### Work Energy Theorem Derivation

According to the equations of motion,

v2 = u2 + 2as

Where,

v = final velocity of the object;

u = initial velocity of the object;

a = constant acceleration; and

s = displacement of the object.

We can also write the above equation as,

v2 - u2 = 2as

Substituting the values of the vector quantities, we get;

v2 - u2 = 2a.d

By multiplying both sides of the equation by m/2, we get:

½ mv2 - ½ mu2 = ma.d

According to Newton’s second law, we know that F= ma,

Hence, the above equation can be written as;

½ mv2 - ½ mu2 = F.d

We also know that W= F.d and, K.E. = (mv²)/2,

This changes the equation to:

Kf – Ki = W

Hence, we have:

ΔK = W

Where, ΔK = Kf – Ki (change in kinetic energy)

This is the derivation of Work-Energy Theorem. Thus, we can say that the work done on an object is equal to the change in the kinetic energy of the object.

### Work Energy Theorem for Variable Force

The force that we come across everyday is usually variable forces. Deriving the work energy formula for variable force is a bit hectic. Let us consider a graph with the variable force in one axis and displacement in the other.

In this graph, x-axis is taken as displacement, and y-axis is force. Let us divide the area of the graph into infinitely small width rectangles along the x-axis.

We can assume that, for an infinitely small displacement Δx, the force applied is constant. So, we get;

ΔW = F(x) Δx

Now we can define work as a definite integral of force over the net displacement as:

W = $\int_{xi}^{xf}$ Fxdx

Since kinetic energy is given by:

K = ½ mv2

The change in kinetic energy with respect to time can be described as,

dK/dt = d/dt(½ mv2)

Or,

dK/dt = mdv/dt.

Newton’s second law of motion states that acceleration is the change in the velocity of an object with respect to time. So, the above equation can be changed as,

dK/dt = mav

Since F= ma, and velocity is the rate of change displacement over time rate, we can write the above equation as,

dK/dt = F.dx/dt

If we cancel out the time derivative from both sides of the equation, we get,

dK = Fdx

Integrating the above equation along the x-axis of the second graph, we get,

$\int_{ki}^{ki}$  dK =  $\int_{xi}^{xf}$ Fxdx

Or, Kf – Ki =  $\int_{xi}^{xf}$ Fxdx

Or we can write,

ΔK = $\int_{xi}^{xf}$ Fxdx

The right-hand side of this equation denotes work done

Hence we get,

ΔK = W

The above equation is the proof of work-energy theorem for the variable force.

### Work Energy Theorem for Constant Force Derivation

Let us consider an object of mass m which is moving under the influence of constant force F. From Newton’s second law of motion:

F = ma

Where,

a = acceleration of the object

The velocity of the object increases from v1 to v2 due to the acceleration, and the object displaces by a distance d.

a = (v22−v12)/2d, or

Now we have,

F = m (v22−v12)/2d, or

Fd = m (v22−v12)/2d, or

Fd = ½ m.v22 – ½ m v12 -------(i)

Fd is the work done by the force F to move the object through a distance d.

In equation (i), the quantity

K2 = m.v22/2, is the final Kinetic energy of the object, and the quantity

K1=mv12/2

Is the initial Kinetic of the object

Thus equation (i) becomes

W=K2-K1=ΔK------(ii)

Where,

ΔK = change in KE of the object.

From equation (ii), it is clear that the work done by a force on an object is equal to the change in kinetic energy of the object.

1. Steps to approach problems on work energy theorem?

Ans- These are the following steps which should be considered while solving problems:

1: Draw the FBD (free body diagram) of the object to identify the forces that are acting on the object.

2: Finding the initial kinetic energy and the final kinetic energy of the object.

3: Equate the values according to the work energy theorem.

2. Consider a body having mass 0.5 kg, and travelling in a straight line. Here, the velocity, “v” is given as x3/2, where a = 5 m-1/2 s-1. Find the work done by the net force during its displacement from: x = 0 to x = 2 m.

Ans- Since, W = Kf - Ki

Putting the value of x = 0 in the equation:

v =a x3/2

We get the initial velocity,

vi=0, and

hence Ki=0,

Similarly to find final velocity vf

by putting x=2,

we get Kf= ½ mv2.

4. What is the work-energy principle?

The work energy principle depicts that, “the rise in the kinetic-energy of a rigid body is because of the +ve work done on the body, by the net resultant force that acts on it”. Conversely, a decrease in kinetic energy is caused by the negative work done by the resultant force on the body.