**Work Energy Theorem** relates the work done and the subsequent change in kinetic energy of the object. To understand the relation between these two quantities, consider an example from our daily lives. Suppose you go for a run. The harder you run, the faster you can go. The speed at which you can run just depends on the effort your muscles are putting. Here we can see that if our muscles work more, the speed at which we can run will be more and hence kinetic energy will be more. This shows that the two quantities are related and is proved by Work-Energy Theorem.

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## What is Work?

Work is said to be done on an object when an external force causes the object to displace from its position. Work is equal to the product of the component of force in the direction of displacement and the magnitude of the displacement of the same object. The SI unit of work is Joule.

## What is Kinetic Energy?

Kinetic energy is the energy possessed by an object by virtue of its motion. It is the energy required by an object to bring it from the state of rest to the state of its required velocity. It is measured in Joule.

\(K = \frac{1}{2}\rm{m}{\rm{v}^2}\)

## What is Work Energy Theorem?

*The work-energy theorem states that the net work done by the external forces on an object is equal to the change in kinetic energy of the object*. It is important to note that the work-energy theorem involves net work done by all forces and not just the work done by a single force. Also, from the above statement, it is clear that work done and kinetic energy are interrelated. Thus,**Kinetic energy** can be defined as the **work** required to accelerate the object from the state of rest to the state of the required velocity.**Work done** can be defined as the **energy** transferred from or to an object by applying a force in the direction of displacement of the object.

## What is the Work Energy Theorem Formula?

As per the statement of the work-energy theorem, if \(∆K\) represents the change in kinetic energy of the body and \(W\) represents the work done on it by the external forces, then:\(∆K = W\)

Or \(K_f – K_i = W\)

Where \((K_f)\) represents the Final Kinetic Energy of the object.

\((K_i)\) represents the Initial Kinetic Energy of the object.

This theorem follows the law of energy conservation, which states that energy can not be created, it can only be transferred from one form to another. To visualize this, consider the example of a box placed on the surface of an inclined plane, as shown below:

The forces acting on the box are weight, normal reaction and friction force. All these are external forces.

According to the work-energy theorem,

\(\rm{Work}\,\rm{done}\,\rm{by}\,\rm{external}\,\rm{forces} = \rm{Change}\,\rm{in}\,\rm{kinetic}\,\rm{energy}\)

If, \(W_g =\) work done by gravity

\(W_N =\) work done by a normal reaction

\( W_f =\) work done by friction force

\(K_f =\) final kinetic energy

\(K_i =\) initial kinetic energy

Then, \(W_g + W_N + W_f = K_f – K_i\)

### Work-done Under a Constant Force

To derive an expression for the work done under a constant force, let us assume that an object with final velocity \(v\), initial velocity \(u\), covers a distance \(s\), under a constant acceleration \(a\). The equation of motion of the object can be given as:

\(v^2 = u^2 + 2as\)

\(v^2 – u^2 = 2as\) —-(i)

By substituting the values of the vector quantities in the equation (i), we get:

\(v^2 – u^2 = 2.\overrightarrow a .\overrightarrow d \) —(ii)

Multiplying both the sides of equation (ii) by \(m/2\), we get:

\(\frac{1}{2}m{v^2} – \frac{1}{2}m{u^2} = \frac{1}{2}m \times 2\left({\vec a.\vec d} \right)\)

\(\frac{1}{2}m\left[{{v^2} – {u^2}} \right] = m(\vec a \cdot \vec d)\) ……..(iii)

From, Newton’s second law of motion, we know that force acting on an object, moving with constant acceleration, can be given as \(\vec F = m\vec a.\)

Using this relation in equation (iii), we get:

\(\frac{1}{2}m\left[{{v^2} – {u^2}} \right] = \vec F.\vec d\) ……..(iv)

Here, \(\frac{1}{2}m\left[{{v^2}}\right]\) represents the Final Kinetic Energy \((K_f)\) of the object.

\(\frac{1}{2}m\left[{{u^2}}\right]\) represents the Initial Kinetic Energy \((K_i)\) of the object.

\(\vec F.\vec d\) is the work done

Thus, equation (iv) can be written as:

\(K_f – K_i = W\)

\(∴ ∆ K = W\)

Where \(∆K\) is the change in kinetic energy.

Thus, the work done by an object is equal to the change in its kinetic energy. It is the proof for the Work-Energy Theorem.

### Work done Under a Variable Force

In our daily lives, a constant force is rarely seen. In fact, a constant force is an ideal scenario, and the forces that occur all around us are all variable. Work energy theorem holds for all forces. Thus, even for a variable force, the work-energy theorem can be proved. However, the procedure may be just a little tricky.

Consider one dimensional case. We can draw the graph between the force and displacement of an object by plotting the force acting on it along one axis and its subsequent displacement along the other axis.

To prove the work-energy theorem, let us take the force \(F\) along the \(y-\)axis and the displacement \(x\) along the \(x-\)axis, as shown in the graph below:

Divide the area under the curve into rectangles of infinitely small width along the \(x-\)axis, as shown in the graph below:

For an infinitely small displacement of “\(∆x\)” of the object, it can be assumed that the force acting on it is constant. Now, during this infinitely small displacement, the work done on the object can be given as:

\(W = F(x)∆x\)

By summing over all the rectangles in the second graph, we come up with the total work. It can be expressed as,

\(W = \sum\limits_ { {x_i}}^ { {x_f}} {F\left( x \right)\Delta x} \)

When the number of rectangles becomes infinite, the width of each rectangle will tend to zero. Thus, we can write the equation of total work done as:

\(W = \underset{ {x \to 0}}{\mathop \rm{lim}} \sum\limits_{{x_i}}^{{x_f}} {F\left( x \right)\Delta x} \)

The total work done by the force is equal to the definite integral of the force over the entire displacement, thus

\(W = \int_{{x_i}}^{{x_f}} {F\left(x \right)\Delta x} \) ……(i)

The kinetic energy of an object of mass \(m\) moving with a velocity \(v\) can be given as:

\(K = \frac{1}{2}m{v^2}\)

Differentiating both sides of the equation with respect to time, we get

\(\frac{{dK}}{{dt}} = \frac{1}{2}\frac{{d\left({m{v^2}} \right)}}{{dt}}\)

\(\frac{{dK}}{{dt}} = \frac{{mvdv}}{{dt}}\)

Acceleration of an object is equal to the rate of change of its velocity; thus, \(a = \frac{{dv}}{{dt}}\) and velocity of an object is equal to the rate of change of displacement, thus, \(v = \frac{{dx}}{{dt}}.\)

Substituting this value,

\(\frac{{dK}}{{dt}} = mv \cdot a\)

\(\frac{{dK}}{{dt}} = \frac{{m \cdot a \cdot dx}}{{dt}}\)

Since force acting on the object \(F = m \cdot a\)

\(\frac{{dK}}{{dt}} = \frac{{F \cdot dx}}{{dt}}\)

Or, \(dK = F \cdot dx\)

Integrating both sides of the equation,

\(\int_{{K_i}}^{{K_f}} {dK = } \int_{{x_f}}^{{x_i}} {F \cdot dx} \)

\(\therefore {K_f} – {K_i} = \int_{{x_f}}^{{x_i}} {F \cdot dx} \)

The left-hand side of the equation represents the change in kinetic energy, therefore:

\(\Delta K = \int_{{x_f}}^{{x_i}} {F \cdot dx} \) ……(ii)

Comparing equations (i) and (ii), we get:

\(∆K = W\)

This gives us, \(\rm{Change}\,\rm{in}\,\rm{kinetic}\,\rm{energy} = \rm{Work}\,\rm{done}\). Thus, the work-energy theorem is proved for a variable force.

## Applications of Work Energy Theorem

Work energy theorem plays an important role in analyzing situations where a rigid body moves under the action of several forces. A rigid body has a rigid lattice structure due to which it cannot store potential energy; thus, it can only possess kinetic energy. Thus, the work done by any force on a rigid body is equal to the change in kinetic energy of the body. This forms the basis of the work-energy equation for rigid bodies.

## Problems Based on Work Energy Theorem

*Q.1. A block of mass \(20\,\rm{kg}\) starts moving up the incline with \(10\,\rm{m/s}\). It reaches the top and comes back to its initial position, and stops. What is the work done by friction in the whole process?*

** Ans:** Forces acting on the block are gravity, normal reaction, and frictional force; all these are external forces, So, Applying the Work-Energy Theorem:

\(W_g + W_n + W_f = K_f – K_i\)

Here, the mass of the box, \(m = 20\,\rm{kg}\)

The velocity of the box, \(v = 10\,\rm{m/s}\)

\({K_i} = \frac{1}{2} \times m \times {v^2} = \frac{1}{2} \times 20 \times 100 = 1000\,{\text{J}}\)

Since the box stops finally, \(K_f = 0\)

\(W_f = 0 – 1000\,\rm{J}\)

\(W_f = – 1000\,\rm{J}\)

Thus, the work done by friction is negative.

*Q.2. A bullet of mass \(10\,\rm{g}\) moving with a velocity of \(500\,\rm{m/s}\), strikes a tree and goes out from the other side with a velocity of \(400\,\rm{m/s}\). Calculate the work done by the bullet (in joules) in passing through the tree.*

** Ans:** Given,

Mass of the bullet, \(m = 10\,\rm{g} = 0.01\,\rm{kg}\)

The initial velocity of the bullet, \(u = 500\,\rm{m/s}\)

The final velocity of the bullet, \(v = 400\,\rm{m/s}\)

Applying the work-energy theorem, the change in kinetic energy of the bullet will be equal to the work done on the bullet by the tree. Thus,

\(W_b = K_f – K_i\)

\({W_b} = ~\frac{1}{2} \times m \times {v^2} – \frac{1}{2} \times m \times {u^2}\)

\({W_b} = \frac{1}{2} \times m \times \left[{{v^2} – {u^2}} \right]\)

\({W_b} = \frac{{0.01}}{2} \times [160000 – 250000]\)

\(W_b = – 450\,\rm{J}\)

Therefore work done by the bullet is \(450\,\rm{J}\).

## Summary

The work-energy theorem states that the net work done by the external forces on an object is equal to the change in kinetic energy of the object. If \(∆K\) represents the change in kinetic energy of the body and \(W\) represents the work done on it by the external forces, then: \(∆K = W\).

## Frequently Asked Questions on Work Energy Theorem

** Q.1. State work-energy theorem.** Work energy theorem states that the work done by net forces on a body is equal to the change in kinetic energy of the body.

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** Q.2. Write the limitation of the Work-energy theorem.** While applying this theorem, we have considered that even large objects behave like particles, which might not be true for all the objects around us.

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** Q.3. How to apply the work-energy theorem?** Step1: Draw FBD of the given object to identify all the forces acting in it.

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Step2: Calculate the change in kinetic energy of the object by subtracting the final kinetic energy from the initial.

Step3: Equate the work done by external forces to the change in kinetic energy.

** Q.4. Define Work.** Work is defined as the energy transferred to/ from an object by applying an external force along with displacement.

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** Q.5. Can we apply the work-energy theorem when the body is moving under variable force?** Yes, we can apply the work-energy theorem for a body moving under variable force. (the derivation has been done above).

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*We hope this detailed article on Work Energy Theorem helps you in your preparation. If you get stuck do let us know in the comments section below and we will get back to you at the earliest.*

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