Q130 / 180 · COMEDK 2024 (Afternoon Shift) · Physics
An object of mass $3\text{ kg}$ moves due to an applied constant force such that its position along X axis is given by $x = \dfrac{t^{\mathrm{2}}}{3}$ where $x$ is in meters and $t$ in seconds. The work done in 1 second is
- A1 J
- B2 J
- C6 J
- Correct answerD$\dfrac{2}{3}$ J
Solution
$x = \dfrac{t^2}{3} \Rightarrow v = \dfrac{dx}{dt} = \dfrac{2t}{3}$
At $t = 0$: $v_i = 0$, at $t = 1$ s: $v_f = \dfrac{2}{3}$ m/s
By work-energy theorem:
$W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = \dfrac{1}{2} \times 3 \times \left(\dfrac{2}{3}\right)^2 - 0 = \dfrac{3}{2} \times \dfrac{4}{9} = \dfrac{2}{3}$ J
At $t = 0$: $v_i = 0$, at $t = 1$ s: $v_f = \dfrac{2}{3}$ m/s
By work-energy theorem:
$W = \dfrac{1}{2}mv_f^2 - \dfrac{1}{2}mv_i^2 = \dfrac{1}{2} \times 3 \times \left(\dfrac{2}{3}\right)^2 - 0 = \dfrac{3}{2} \times \dfrac{4}{9} = \dfrac{2}{3}$ J
Hence, the correct answer is option: 'D'.