Question

Under the action of a force, a $2lg$ body moves such that it position $x$ as a function of time is given by $x=\frac{t^3}{3}$, where $x$ is in metre and $t$ in seconds. The work done by the force in the first two seconds is

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

Answer 1

The correct option is C $16J$

Given : Mass m$=2kg$ $x\left(t\right)=\frac{t^3}{3}$

Time $t=2s$

Differentiating position $x\left(t\right)=velocityv$

$=\frac{d}{dt}\left(\frac{t^3}{3}\right)=\frac{1}{3}.3t^2=t^2$

At $t=2s\Rightarrow v=t^2=4m/s$

Kinetic energy $=\frac{1}{2}m{v}^2=16J$