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Q.

A body of mass 3 kg is under a force, which causes a displacement in it is given by $S = \frac{t^{3}}{3}$ (in m). Find the work done by the force in first 2 seconds

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a

3.8 J

b

2 J

c

24 J

d

5.2 J

answer is D.

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Detailed Solution

$S = \frac{t^{3}}{3} ∴ dS = t^{2} dt$

$a = \frac{d^{2} S}{{dt}^{2}} = \frac{d^{2}}{{dt}^{2}} [\frac{t^{3}}{3}] = 2 tm / s^{2}$

Now work done by the force

$W = \int_{0}^{2} F \cdot dS = \int_{0}^{2} ma \cdot dS$

$\int_{0}^{2} 3 \times 2 t \times t^{2} dt = \int_{0}^{2} 6 t^{3} dt = \frac{3}{2} {[t^{4}]}_{0}^{2} = 24 J$

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A body of mass 3 kg is under a force, which causes a displacement in it is given by S=t33 (in m). Find the work done by the force in first 2 seconds