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A body of mass $3 \mathrm{~kg}$ is under a constant force which causes a displacement $s$ in metres in it, given by the relation $s=\frac{1}{3} t^2$, where $t$ is in seconds. Work done by the force in 2 seconds is:
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Verified Answer
The correct answer is:
$\frac{8}{3} \mathrm{~J}$
It is given that:
$s=\frac{t^2}{3}$
Differentiating w.r.t. $t$ we have
$U=\frac{d s}{d t}=\frac{2 t}{3}$
Again differentiating w.r.t. $t$ we have
$a=\frac{d^2 s}{d t^2}=\frac{2}{3}$
Now $\mathrm{W}=\int \mathrm{F} d s=\int \mathrm{mads}$
$\begin{aligned}
& =\int m \times \frac{2}{3} \times \frac{2 t}{3} d t \\
& =\frac{4}{9} m \int t d t \\
W & =\frac{4}{9} m \times\left[\frac{t^2}{2}\right]_0^2 \\
& =\frac{4}{9} \times 3 \times 2=\frac{8}{3} \mathrm{~J}
\end{aligned}$
$s=\frac{t^2}{3}$
Differentiating w.r.t. $t$ we have
$U=\frac{d s}{d t}=\frac{2 t}{3}$
Again differentiating w.r.t. $t$ we have
$a=\frac{d^2 s}{d t^2}=\frac{2}{3}$
Now $\mathrm{W}=\int \mathrm{F} d s=\int \mathrm{mads}$
$\begin{aligned}
& =\int m \times \frac{2}{3} \times \frac{2 t}{3} d t \\
& =\frac{4}{9} m \int t d t \\
W & =\frac{4}{9} m \times\left[\frac{t^2}{2}\right]_0^2 \\
& =\frac{4}{9} \times 3 \times 2=\frac{8}{3} \mathrm{~J}
\end{aligned}$
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