Q103 / 200 · NEET 2006 · Physics

A body of mass $3 \mathrm{~kg}$ 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:

  • A
    $\frac{19}{5} \mathrm{~J}$
  • B
    $\frac{5}{19} \mathrm{~J}$
  • C
    $\frac{3}{8} \mathrm{~J}$
  • Correct answer
    D
    $\frac{8}{3} \mathrm{~J}$

Solution

It is given that: $s=\frac{t^2}{3}$ Differentiating w.r.t. $t$ we have $U=\frac{d s}{d t}=\frac{2 t}{3}$ Again differentiating w.r.t. $t$ we have $a=\frac{d^2 s}{d t^2}=\frac{2}{3}$ Now $\mathrm{W}=\int \mathrm{F} d s=\int \mathrm{mads}$ $\begin{aligned} & =\int m \times \frac{2}{3} \times \frac{2 t}{3} d t \\ & =\frac{4}{9} m \int t d t \\ W & =\frac{4}{9} m \times\left[\frac{t^2}{2}\right]_0^2 \\ & =\frac{4}{9} \times 3 \times 2=\frac{8}{3} \mathrm{~J} \end{aligned}$

Hence, the correct answer is option: 'D'.

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