Question

A body of mass $m$ is moving with a uniform velocity $u$. A force is applied on the body due to which its velocity changes from $u$ to $v$. How much work is being done by the force?

• A. $\frac{1}{2}m(v^2-u^2)$
• B. $\frac{1}{2}m\left(v^2\right)$
• C. $\frac{1}{2}m(v^2+u^2)$
• D. $\frac{1}{2}m\left(u^2\right)$

Answer 1

The correct option is A $\frac{1}{2}m(v^2-u^2)$

Here we denote work by $W$ and force by $F$ and displacement by $s$ and acceleration by $a$.
Using third equation of motion, $v^2-u^2=2as$

$\therefore a=\frac{(v^2-u^2)}{2s}$

We know $F=ma$

$⟹$ $F=\frac{m(v^2-u^2)}{2s}$

Work done $W=Fs$
Thus $W=\frac{m(v^2-u^2)s}{2s}$
$W=\frac{1}{2}m(v^2-u^2)$.
This is the work done.