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A body of mass $3 \mathrm{~kg}$ is moving under the action of a force which causes a displacement of $\left(\frac{t^3}{3}\right) \mathrm{m}$, where ' $\mathrm{t}$ ' is time in seconds. The work done by the force in first 2 seconds is
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Verified Answer
The correct answer is:
$24 \mathrm{~J}$
Mass of body, $\mathrm{m}=3 \mathrm{~kg}$
Displacement $\mathrm{x}=\frac{\mathrm{t}^3}{3} \mathrm{~m}$
Velocity, $v=\frac{d x}{d t}$
$$
=\mathrm{t}^2
$$
from work energy theorem
$$
\begin{aligned}
& W=K E_f-K E_i \\
& =\frac{1}{2} m v^2-0 \\
& =\frac{1}{2} m\left(t^2\right)^2=\frac{1}{2} m t^4=\frac{1}{2} \times 3 \times(2)^4=24 J
\end{aligned}
$$
Displacement $\mathrm{x}=\frac{\mathrm{t}^3}{3} \mathrm{~m}$
Velocity, $v=\frac{d x}{d t}$
$$
=\mathrm{t}^2
$$
from work energy theorem
$$
\begin{aligned}
& W=K E_f-K E_i \\
& =\frac{1}{2} m v^2-0 \\
& =\frac{1}{2} m\left(t^2\right)^2=\frac{1}{2} m t^4=\frac{1}{2} \times 3 \times(2)^4=24 J
\end{aligned}
$$
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