Q134 / 160 · AP EAMCET 2022 (08 Jul Shift 1) · Physics

The work done in stretching a spring of natural length 25 $\mathrm{cm}$ and spring constant $50 \mathrm{Nm}^{-1}$ from $50 \mathrm{~cm}$ to $60 \mathrm{~cm}$ is

  • Correct answer
    A
    $1.5 \mathrm{~J}$
  • B
    $2 \mathrm{~J}$
  • C
    $3.5 \mathrm{~J}$
  • D
    $5 \mathrm{~J}$

Solution

$\Delta \mathrm{x}_{\mathrm{i}}=(50-25) \mathrm{cm}=25 \mathrm{~cm}$ $\Delta \mathrm{x}_{\mathrm{f}}=(60-25) \mathrm{cm}=35 \mathrm{~cm}$ So, work done $=\Delta U$ $\begin{aligned} & =\frac{1}{2} \mathrm{~K}\left[\Delta \mathrm{x}_{\mathrm{f}}^2-\Delta \mathrm{x}_{\mathrm{i}}^2\right] \\ & =\frac{1}{2} \times 50\left[35^2-25^2\right] \times 10^{-4} \\ & =1.5 \mathrm{~J}\end{aligned}$

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

Relevant Questions