Q130 / 160 · TS EAMCET 2020 (11 Sep Shift 2) · Physics

Under action of force, a $2 \mathrm{~kg}$ body moves such that its position $x$ as function of time $t$ is given by $x=\alpha t^2 / 2$, where $x$ is in metre, $t$ is in seconds and $\alpha=1 \mathrm{~m} / \mathrm{s}^2$. The work done by the force in the first two seconds is

  • Correct answer
    A
    $4 \mathrm{~J}$
  • B
    $16 \mathrm{~J}$
  • C
    $40 \mathrm{~J}$
  • D
    $2 \mathrm{~J}$

Solution

Given, $x=\frac{\alpha t^2}{2}$ As we know that, $v=\frac{d x}{d t}=\frac{d}{d t}\left(\frac{\alpha t^2}{2}\right)=\alpha t$ $\ldots(i)$ According to work-energy theorem Work done, $W =\frac{1}{2} m v^2=\frac{1}{2} m(\alpha t)^2 \quad \text { [using Eq. (i)] }$ $frac{1}{2} m \alpha^2 t^2=\frac{1}{2} \times 2 \times(1)^2 \times(2)^2=4 \mathrm{~J}$

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

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