Question

A body of mass $2kg$ moving under a force has relation between displacement $x$ and time $t$ as $x=\frac{t^3}{3}$ where $x$ is in metre and $t$ is in sec. The work done by the body in first two second will be

• A. $1.6joule$
• B. $16joule$
• C. $160joule$
• D. $1600joule$

Answer 1

The correct option is B $16joule$

work done = change in kinetic energy

$x=\frac{t^3}{3}$

velocity $v=\frac{dx}{dt}=\frac{d\left(\frac{t^3}{3}\right)}{dt}=t^2$

given, $t=2s$

Velocity $=t^2=2^2=4$

$W=\frac{1}{2}m{v}^2$

$=\frac{1}{2}\times 2\times 4^2$

$=16J$