# 5.1.1 Work-Energy Theorem

Consider the rectilinear motion of a mass m being pushed by a constant force F, causing the mass’s speed to increase from u to v over a distance of Δs.

Since a constant F results in constant acceleration a, we can apply the equation of motion.

\begin{aligned}{{v}^{2}}&={{u}^{2}}+2a\Delta s\\2a\Delta s&={{v}^{2}}-{{u}^{2}}\\ma\Delta s&=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}\end{aligned}

But ma is F, so

$F\Delta s=\frac{1}{2}m{{v}^{2}}-\frac{1}{2}m{{u}^{2}}$

Obviously, the quantity on the RHS represents the increase in kinetic energy of the mass.

The quantity on the LHS is given the name work. It is usually denoted by the symbol W and has the unit J.

This leads us to the work-energy theorem,

$W=F\Delta s=\Delta KE$

Recognize the resemblance with the impulse-momentum theorem

$J=F\Delta t=\Delta p$

Basically, a force can change the KE and momentum of a body. If we multiply F with the Δs of the body, we obtain the work done by the force, which can be equated to changes in the body’s KE. On the other hand, if we multiply F with Δt, we obtain the impulse of the force, which can be equated to changes in the body’s momentum.

Demonstration

Blow Pipe

Concept Test

0804

Video Explanation

Derivation of the KE Formula

Area Under the Force-Displacement Graph