Work Energy Theorem

The Work-Energy Theorem states that the net work done by all the forces acting on a body equals the change in its kinetic energy.

• It establishes a relationship between work and kinetic energy and is one of the most important theorems in mechanics.
• If positive work is done on an object, its kinetic energy increases, while if negative work is done, its kinetic energy decreases.

Formula

W=\triangle KE or W= \frac{1}{2}mv^2_f-\frac{1}{2}mv^2_i

• W = net work done
• m = mass of the body
• v_i​ = initial velocity
• v_f​ = final velocity
• KE_i​ = initial kinetic energy
• KE_f​ = final kinetic energy

Derivation of Work-Energy Theorem for Constant Force

Consider a particle of mass (m) moving along a straight line under the action of a constant force (F). Let:

• Initial velocity = (v_i)
• Final velocity = (v_f)
• Displacement = (d)
• Acceleration = (a)

The force is assumed to be constant in magnitude and direction and acts parallel to the displacement of the particle.

According to Newton’s second law of motion:

F = ma

Using the third equation of motion:

v_f^2 = v_i^2 + 2ad

Rearranging the equation for displacement:

d = \frac{v_f^2-v_i^2}{2a}

The work done by the force is:

W = Fd

Substituting the values of (F) and (d):

W = ma\left(\frac{v_f^2-v_i^2}{2a}\right)

Cancelling (a):

W = \frac{m(v_f^2-v_i^2)}{2}

Separating the terms:

W = \frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2

Since kinetic energy is given by:

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

Therefore:

W = KE_f - KE_i

Hence,

W = ΔKE

Thus, the net work done on a particle is equal to the change in its kinetic energy.

Derivation of Work-Energy Theorem for Variable Force

When a variable force F(x) acts on a particle over a small displacement dx, the small work done is:

dW = F(x)dx

Hence, total work done from ( x_i ) to ( x_f ) is:

W = \int_{x_i}^{x_f}F(x)dx

The kinetic energy of a particle is:

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

Differentiating with respect to time:

\frac{dK}{dt} = mv\frac{dv}{dt}

Using:

F = ma

and

v = \frac{dx}{dt}

we get:

dK = Fdx

Integrating both sides:

\int dK = \int F(x)dx

Therefore:

K_f-K_i = \int_{x_i}^{x_f}F(x)dx

Since the integral of force over displacement is work done:

W = ΔK

Hence, the net work done by a variable force is equal to the change in kinetic energy.

Sample Problem

Problem 1: A 2 kg ball is dropped from a height of 10 m. Calculate the work that has been done on the ball between the moment it is released and the moment when it hits the ground. Assume that air resistance can be neglected.

Solution: Given, 

Mass of the ball, m=2 kg.

Initial height of the ball, h_i =10 m.

Final height of the ball, h_f =0 m.

We must calculate the amount of work done on the ball as it hits the ground.

The ball is dropping freely here, conserving energy. We know that the difference in kinetic energy equals to the work done.

Because the ball is in a motionless condition when it is dropped, it has no kinetic energy at first.

All the ball's potential energy is transformed to kinetic energy when the ball touches the ground.

So, Potential Energy of Ball at height h_i is,

E_p = mgh

 E_p=2\times 9.8 \times 10=196 J

Now, The ball had 196 J of potential energy when it was released and 0 J of kinetic energy. 

When the ball hit the ground, it had 0 J of potential energy and 196 J of kinetic energy. 

Therefore, Initial Kinetic Energy KE_i=0 J and KE_f=196 J.

By work-energy theorem:

W_net = ΔKE

KE_f − KE_i = 196−0 

= 196 J

Therefore, The net work done on the ball is 196 J.

Problem 2: A car of mass 500 kg travelling at a speed of 16 m/s applies the car's brakes at some point. The car's brakes provide a frictional force of 4000 N. Determine the stopping distance of the car.

Solution: Given, 

Mass of the car, m = 500 kg

Speed of the car, v = 16 m/s

Frictional force of brakes, \vec{F}  = 4000 N

Applying the work-energy theorem as:

The change in the kinetic energy is equal to the work done by the frictional force of the car's brakes.

Therefore, Initial Kinetic energy of car at the moment brakes were applied is,

KE_i = \frac{1}{2}mv^2

KE_i = \frac{1}{2}\times500\times (16)^2  ⇒ 64 kJ.

As the car is at rest, Final Kinetic energy =0, therefore \triangle KE  = 64 kJ.  ......(1)

Consider the displacement of car is ΔX. Since the direction of the applied force and the displacement are in opposite directions, θ=180°. 

Then the work done is given by:

W=F\triangle x cos\theta

W = (4000)(\triangle x)(\cos{180\degree})

 ......(2)                                                                                                                                                      

By work-energy theorem, The change in kinetic energy is equal to the work done.

64000 = -4000 \triangle x

\triangle x = 16m

Therefore, The stopping distance of car is 16 m.

Problem 3: The net propulsion force of a 2000 kg rocket is 600 N. The rocket accelerates uniformly from an initial velocity of 50 m/s to a final velocity of 80 m/s in a short amount of time. Assume that the rocket's mass remains constant while it burns fuel and that the net force is directed in the direction of motion. How much net work has been done on the rocket (in KJ)?

Solution: Given,

Mass of rocket, m = 2000 kg.

Net propulsion Force, F = 600 N.

Initial Velocity of rocket, v_i = 50 m/s.

Final Velocity of rocket, v_f = 80 m/s.

 Applying Work Energy Theorem, 

W = ΔKE

Initial Kinetic energy of the rocket KE_i is given by:

 KE_i=\frac{1}{2}mv_i^2

KE_i= 0.5(2000kg)(50)^2=2500000J

Similarly, Final Kinetic energy of rocket KE_f is given by,

 KE_f=\frac{1}{2}mv_f^2

KE_f= 0.5(2000kg)(80)^2=6400000J

Net Work done = The change in the kinetic energy at the two velocities:

W = ΔKE

6400000−2500000 =3 900000 J

Therefore, Net work done on the rocket is 3900 kJ.

Problem 4: A 100 g arrow is fired from a bow whose string exerts an average force of 150 N on the arrow over a distance of 50 cm. What is the speed of the arrow as it leaves the bow?

Solution: Given, 

Mass of arrow, m = 100g

Force exerted on the arrow, F = 150N

Displacement of string, d = 50cm

Applying Work-Energy Theorem,

W = ΔKE

W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2

Work is equal to the force times the displacement over which the force acted.

W = Fd

Therefore, Fd=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2

Substituting the values, we get

 (150N)(0.5m)=\frac{1}{2}(0.01kg)(v_f)2−\frac{1}{2}(0.01kg)(0m/s)^2               

v_f^2=15000m/s

v_f=122.47m/s

Therefore, Velocity of arrow as it leaves the bow is 122.47 m/s.

Problem 5: Ronaldo kicked a football, and it traveled 50m with a constant velocity, how much work does Ronaldo do? 

Solution:  Work is the product of a net force over a given distance.

W = F_net d

There must be a net force exerted and a non-zero displacement of the object in order for it to do work.

The displacement is provided, but we must solve for the force.

According to Newton's Second Law of Motion, the football moves at a constant velocity, which means its acceleration is zero.

a=\frac{v_f−v_i}{t}

Since, v_f=v_i, a=0 m/s²

 F = ma 

= m(0) 

= 0N

If the force is zero, then the work is also zero. Therefore, Work done, W is 0 J.

Problem 6: A 20 kg box is moving at 4 m/s. A net force of 100 N is constantly applied on the box in the direction of its movement until it has moved 25 m. What is the approximate final velocity of the box?

Solution:  Given, 

Mass of box, m = 20kg

Initial Velocity of box, v_i = 4 m/s

Force acting on the box, F = 100N.

Displacement of the box, d = 25m.

Applying Work-Energy Theorem,

W = ΔKE

 W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2

Work done by the block, W = Fd

W = 100 × 25 J 

= 2500 J

Initial Kinetic Energy, KE_i=\frac{1}{2}mv_i^2

KE_i=\frac{1}{2}(20)(4 m/s)^2

KE_i=160 J

Substituting the value, we get,

 2500 J=\frac{1}{2}(20)v_f^2-160 J 

v_f^2=532m/s

 v_f=23.05m/s

Therefore, appropriate final velocity is 23.05 m/s.

Unsolved Problem

Question 1: A 1.5 kg ball is thrown vertically upward with a speed of 10 m/s. Calculate the net work done on the ball by gravity when it reaches the highest point.

Question 2: A 5 kg box is pushed across a friction less horizontal floor with a constant force of 20 N over a distance of 4 m. Find the work done on the box.

Question 3: A 1000 kg car moving at 25 m/s brakes uniformly and comes to rest. The brakes provide a variable frictional force that depends on distance F(x) = 4000 + 10x N, where x is in meters. Find the distance the car travels before stopping using work-energy theorem.

Question 4: A 3 kg block is projected up a rough incline with speed 8 m/s. The work done by friction along the incline is 12 J. Find the maximum height reached by the block.

Question 5: A 500 kg elevator starts from rest and is lifted vertically by a motor. The motor does 1.5 × 10⁵ J of work on the elevator. Find the velocity of the elevator after being lifted 10 m, assuming gravity acts downward.

Question 6: A particle moves along a straight line. The v²–x graph of the particle is a straight line starting from v² = 4 m²/s² at x = 0 and reaching v²=20 m²/s² at x=8 m. Using the graph, find the acceleration of the particle and its velocity at x=8 m.
(Hint: Slope of v²–x graph = 2a.)