Question

A body of mass m starts moving from rest along x-axis so that its velocity varies as $v = a \, \sqrt{s}$ where a is a constant and s is the distance covered by the body. The total work done by all the forces acting on the body in the first t seconds after the start of the motion is :

• A. $\frac{1}{8} m \, a^4 \, t^2$
• B. $8 \, m \, a^4 \, t^2$
• C. $4 \, m \, a^4 \, t^2$
• D. $\frac{1}{4} \, m \, a^4 \, t^2$

Answer 1

Answer: (A)

$\frac{1}{8} m \, a^4 \, t^2$

Answer 2

Velocity of the body is given by
$v=a \sqrt{s}$
Differentiating w.r.t. $t$, we get
Acceleration, $a'=\frac{d v}{d t}=a \frac{1}{2} s^{-1 / 2} \cdot \frac{d s}{d t}=a \frac{1}{2 \sqrt{s}} \cdot v$ $\Rightarrow a'=\frac{a}{2 \sqrt{s}} \cdot a \sqrt{s}=\frac{a^{2}}{2}$
Force on the body is $F=m a'=\frac{m a^{2}}{2}$
Distance covered by the body is given by
$s=u t+\frac{1}{2} a' t^{2}$
$\Rightarrow s=\frac{1}{2} \cdot \frac{a^{2}}{2} t^{2}$
Work done =Force $\times$ Distance $=\frac{m a^{2}}{2} \cdot \frac{1}{2} \frac{a^{2}}{2} t^{2}=\frac{1}{8} m a^{4} t^{2}$