Q128 / 160 · AP EAMCET 2024 (18 May Shift 1) · Physics

A spring of $5 \times 10^3 \mathrm{Nm}^{-1}$ spring constant is stretched initially by 10 cm from unstretched position. The work required to stretch it further by another 10 cm is

  • Correct answer
    A
    $75 \mathrm{~N}-\mathrm{m}$
  • B
    $50 \mathrm{~N}-\mathrm{m}$
  • C
    $76 \mathrm{~N}-\mathrm{m}$
  • D
    $82 \mathrm{~N}-\mathrm{m}$

Solution

The work done required to stretch a spring from $\mathrm{x}_1$ to $x_2$ is $\mathrm{W}=\frac{1}{2} \mathrm{k}\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)$ $=\frac{1}{2} \times 5 \times 10^3\left[\left(20 \times 10^{-2}\right)^2-\left(10 \times 10^{-2}\right)^2\right]=75 \mathrm{~N}-\mathrm{m}$

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

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