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(D) Given: Mass $m = 3\, kg$,displacement $s = \frac{1}{3}t^2$.First,find the velocity $v$ by differentiating $s$ with respect to $t$:$v = \frac{ds}{d

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$\frac{8}{3}\, J$

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(D) Given: Mass $m = 3\, kg$,displacement $s = \frac{1}{3}t^2$.First,find the velocity $v$ by differentiating $s$ with respect to $t$:$v = \frac{ds}{dt} = \frac{d}{dt}(\frac{1}{3}t^2) = \frac{2}{3}t$.Next,find the acceleration $a$ by differentiating $v$ with respect to $t$:$a = \frac{dv}{dt} = \frac{d}{dt}(\frac{2}{3}t) = \frac{2}{3}\, m/s^2$.The force $F$ acting on the body is $F = m 	imes a = 3 	imes \frac{2}{3} = 2\, N$.The work done $W$ is given by $W = \int F \, ds$. Since $ds = v \, dt$,we have $W = \int_0^2 F \cdot v \, dt$.$W = \int_0^2 2 \cdot (\frac{2}{3}t) \, dt = \frac{4}{3} \int_0^2 t \, dt$.$W = \frac{4}{3} [\frac{t^2}{2}]_0^2 = \frac{4}{3} 	imes \frac{4}{2} = \frac{8}{3}\, J$.

$A$ body of mass $3\, kg$ 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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