Question

A body starts from rest and acquires a velocity $V$ in time $T$. The work done on the body in time $t$ will be proportional to

• A. $\frac{V}{T}t$
• B. $\frac{V^2{t}^2}{T}$
• C. $\frac{V^2}{T^2}t$
• D. $\frac{V^2}{T^2}{t}^2$

Answer 1

The correct option is D $\frac{V^2}{T^2}{t}^2$

$v=0+at\Rightarrow a=\frac{V}{T}$
velocity of the body after time $t$ is
$v=0+at\Rightarrow v=\frac{V}{T}t$
Let the mass of the body be $m$
From work energy theorem, the work done is change in kinetic energy.
$W=\frac{1}{2}m{v}^2-0=\frac{m{V}^2{t}^2}{2T^2}$
Hence
$W\propto \frac{V^2}{T^2}{t}^2$