Last updated on June 14th, 2021 at 04:18 pm

**Work-Kinetic Energy Theorem with derivation**: In this post, we will discuss the special relationship between work done on an object and the resulting kinetic energy of the object and come up with the *statement of the work-kinetic energy theorem*. We will also see **how to derive the equation of the work-kinetic energy theorem**.

Everyday experience supports this theorem. If we see a football at rest on the ground and a moment later we see it hurtling through the air, we would conclude that someone did work on the ball, by exerting a large force over a short distance, to make it move.

This correct conclusion illustrates how doing work on an object gives the object increased velocity or kinetic energy.

## State work-kinetic energy theorem

**The special relationship between doing work on an object and the resulting kinetic energy of the object is called the work-kinetic energy theorem.**

## Derivation of the mathematical expression for work-kinetic energy theorem | equation derivation

To develop a mathematical expression that relates work to the kinetic energy, we’ll assume that all of the work done on a system gives the system kinetic energy only.

We will start with the definition of work and then apply Newton’s second law.

To avoid dealing with advanced mathematics, we are assuming that a constant force gives the system a constant acceleration so that you can use the equations of motion.

Since work and kinetic energy are scalar quantities, vector notation will be omitted from the derivation. This is valid as long as the directions of the force and displacement are parallel and the object is moving in a straight line.

Let’s assume that the force is constant and write the equation

for work. W = FΔd …………(1)

Using Newton’s second law, **F** = m**a**

Assume the force and the acceleration are parallel to the direction of the displacement and motion is in one direction. Then omit vector symbols and use F = ma ……. (2)

Let’s Substitute ma for F in the equation for work. W = maΔd…..(3)

Now we will use the equation of acceleration for uniformly accelerated motion. a = (v_{2} − v_{1})/Δt ……… (4)

From equation (3) and (4)

W = m [(v2 − v1)/Δt] Δd

=> W = m (v_{2} − v_{1}) [Δd/Δt] ………………. (5)

Now the equation for displacement for uniformly accelerated

motion. Δd = [(v_{1} + v_{2})/2] Δt …………………(6)

Now, Divide both sides of the equation (6) by Δt to obtain

an equation for **Δd/Δt**.

Δd/Δt = (v_{1} + v_{2})/2…………………..(7)

Rewrite the equation (5) for work by substituting the value for Δd/Δt.

W = m(v_{2} − v_{1})(v_{1} + v_{2})/2**W = (1/2)m(v _{2}^{2} − v_{1}^{2})**………………..(8)

This makes us conclude that **work-done** on an object results in a **change in the kinetic energy** of the object.

W = E_{k}2 – E_{k}1**W = ΔE _{k}** (mathematical expression or equation for the work-kinetic energy theorem is derived here). ………………. (9)

This expression **for the work-kinetic energy theorem** above shows the special relationship between work done on an object and the resulting kinetic energy of the object.

## Summary | Take Away | Suggested reading

So in this post, we have covered the statement of **the work-kinetic energy theorem** and also derived its expression or equation.

Now for the related study, we suggest the following posts for our readers.**Derivation of Kinetic energy equation****Law of conservation of energy**

**Relationship between momentum and Kinetic energy – equation derivation**