Question

State and derive work energy theorem

Answer 1

Step 1. State work energy theorem

Work energy theorem states that the change in kinetic energy of an object is equal to the net work done on it by the net force.

Step 2. Derivation of work energy theorem

Let $\overrightarrow{F}$ is the force applied on a body of mass $M$ to displace it a small distance $d\overrightarrow{s}$ along the direction of the force.

Then the work done by the force in displacing the object is

$dw=\overrightarrow{F}·d\overrightarrow{s}.......\left(i\right)$

Also from Newtons second law of motion we get

$F=Ma......\left(ii\right)$

Where $F$ is the force, $M$ mass and $a$ acceleration

From $\left(i\right)$ and $\left(ii\right)$ we get

$$\begin{gathered} dw=Ma·d\overrightarrow{s} \\ \Rightarrow dw=M\frac{dV}{dt}·d\overrightarrow{s}\left[\because a=\frac{dV}{dt}\right] \end{gathered}$$

Now, work done by the force in order to increase its velocity from $u$(initial velocity) to $v$ (final velocity) is given by

$$\begin{gathered} \Rightarrow \int dw={\int }_u^{v}MVdV\left[\because \frac{ds}{dt}=V\right] \\ \Rightarrow W=M{\left[\frac{V^2}{2}\right]}_u^v \\ \Rightarrow W=\frac{1}{2}M\left[v^2-u^2\right] \\ \Rightarrow W=\frac{1}{2}M{v}^2-\frac{1}{2}M{u}^2 \end{gathered}$$

We know that K.E$=\frac{1}{2}\left(\mathrm{Mass}\right)·{\left(\mathrm{Velocity}\right)}^2$

$\therefore \frac{1}{2}M{v}^2-\frac{1}{2}M{u}^2$ is the rate of change of kinetic energy.

Therefore work done by a force is the rate of change of Kinetic energy