Q127 / 160 · TS EAMCET 2023 (12 May Shift 1) · Physics

The displacement 's' of a body of mass $3 \mathrm{~kg}$ under the action of a force is given by $s=\frac{t^3}{3}$, where ' $s$ ' is in metres and ' $t$ ' is in seconds. The work done by the force in the first two seconds is

  • A
    $32J$
  • B
    $3.8J$
  • C
    $5.2J$
  • Correct answer
    D
    $24J$

Solution

Masses of body, $\mathrm{m}=3 \mathrm{~kg}$ The action of force is given, $s=\frac{t^3}{3}$ Velocity, $v=\frac{\mathrm{ds}}{\mathrm{dt}}=\mathrm{t}^2$ acceleration, $\mathrm{a}=\frac{\mathrm{d} v}{\mathrm{dt}}=2 \mathrm{t}$ $\begin{aligned} & \text { Work done }=\int_0^2 \mathrm{~F} \cdot \mathrm{ds}=\int_0^2 \mathrm{ma} \mathrm{ds} \\ & =\int_0^2 3 \times 2 \mathrm{t} \times \mathrm{t}^2 \mathrm{dt} \\ & =6 \int_0^2 \mathrm{t}^3 \mathrm{dt}=6 \times \frac{\mathrm{t}^4}{4} \\ & =\frac{3}{2}\left(\mathrm{t}^4\right)_0^2=\frac{3}{2} \times(2)^4=24 \mathrm{~J} \end{aligned}$

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

Relevant Questions