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A body of mass 3 kg under constant force which causes a displacement $s$ in meters in it given by $s=\large\frac{1}{3}$$t^2 work done by the force in 2 seconds is $(a)\;\frac{8}{3}\;J\quad (b)\;\frac{19}{5}\;J \quad (c)\; \frac{5}{19}\;J \quad (d)\;\frac{3}{8}\;J$ Can you answer this question? 1 Answer 0 votes s=\large\frac{1}{3}$$t^2=>\large\frac{ds}{dt}=\frac{2}{3}$$t acceleration =a=\large\frac{d^2s}{dt^2} \qquad=\large\frac{2}{3}$$m/s^2$
Force $=m \times a$
$\qquad=3 \times \large\frac{2}{3}$$=2N ds=\large\frac{2}{3}$$tdt$
Work done $= \int F ds$
$W=F \int \limits_0^t \large\frac{2}{3} $$t dt \quad=2 \times \large\frac{2}{3}$$ \times \large \frac{t^2}{2} \bigg|_0^2$
$\quad=\large\frac{4}{3} \times \frac{4}{2}$
$\quad=\large\frac{8}{3}$$\;J$
Hence a is the correct answer.

edited Feb 17, 2014 by meena.p

+1 vote