Question

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

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

Answer 1

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

Acceleration: $a=\frac{V}{T}$

Velocity at time t: $v=at=\frac{Vt}{T}$

Change in KE: $\Delta K=\frac{1}{2}m{v}^2-0=\frac{1}{2}m(\frac{Vt}{T}{)}^2=\frac{m{V}^2{t}^2}{2T^2}$

Work done on the body is equal to change in its kinetic energy.

$\therefore W=\frac{m{V}^2{t}^2}{2T^2}$

$\Rightarrow W\propto \frac{V^2{t}^2}{T^2}$