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Q.

A body of mass 3kg is under a constant force which causes a displacement s in meters 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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a

$\frac{19}{5} J$

b

$\frac{3}{8} J$

c

$\frac{5}{19} J$

d

$\frac{8}{3} J$

answer is C.

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Detailed Solution

$v = \frac{ds}{dt} = \frac{1}{3} (2 t) = \frac{2}{3} t$
Now from work-energy theorem $W = ΔKE$
We have $W = \frac{1}{2} {mv}_{2}^{2} - \frac{1}{2} {mv}_{1}^{2}$
$= \frac{1}{2} (3) {(\frac{2}{3} \times 2)}^{2} - \frac{1}{2} (3) {(\frac{2}{3} \times 0)}^{2} = \frac{8}{3} J$

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A body of mass 3kg is under a constant force which causes a displacement s in meters in it, given by the relation s=13t2, where t is in seconds. Work done by the force in 2 seconds is :