[Physics Class Notes] on Derivation of Work Energy Theorem Pdf for Exam

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.

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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.

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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 the “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.

It is a known fact that we all require Energy in order to Work. 

We Work every day, some people more than others. However, no matter how much or how little one Works, Energy is required by everyone. This is where we come to the Work-Energy Theorem. 

Now here we discuss what Work actually is – When there is a force on a body or an object, that makes the said object or body movement. The direction is not always similar in all of the instances. In fact, it is not given that the direction of the force, as well as the direction of the body or object after the force is applied, will be the same. 

And so, it can be said that Work is done when some force is applied to a certain body or object, and the said body or object changes its position. This also means that they move and cover a certain amount of distance depending on the force that is applied. 

The phenomenon also establishes a relationship that is already there between Kinetic Energy and Work. This relationship is basically called the Work-Energy Theorem. 

This Theorem is very big in terms of its significance to the students. They must not only know its formulae but also its very derivation. That will explain the Workings of a Work-Energy Theorem in a simple manner that is easily understandable. 

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 the 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 every day 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}^{kf}]  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 the 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.

v22−v12=2ad, or

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.