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Q.
A body of mass $6\,kg$ is acted upon by a force which causes a displacement in it given by $x=\frac{t^{2}}{4}$ metre where $t$ is the time in second. The work done by the force in $2$ seconds is
From the question, we see that the mass of the body is $6\, kg$ and the displacement is given by $x=\frac{t^{2}}{4}$.
So, the velocity would be equal to
$v=\frac{d x}{d t}=\frac{t}{2}$.
Thus, $v(0)=0$ and $v(2)=\frac{2}{2}=1$
Hence the initial kinetic energy would be equal to
$KE _{\text {ini }}=\frac{1}{2} m[v(0)]^{2}=0$
and similarly the final kinetic energy would be equal to
$KE _{ fin }=\frac{1}{2} m[v(2)]^{2}=\frac{1}{2} \times 6 \times 1^{2}=3 .$
So, we see that the work done by the force is equal to
$W=K E_{f i n}-K E_{\text {ini }}=3-0=3\, J$.