Q138 / 180 · AIIMS 2014 · Physics

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

  • A
    $6.25 \mathrm{~N} \mathrm{~m}$
  • B
    $1250 \mathrm{~N} \mathrm{~m}$
  • Correct answer
    C
    $18.75 \mathrm{~N} \mathrm{~m}$
  • D
    $25.00 \mathrm{~N} \mathrm{~m}$

Solution

$U_1=\frac{1}{2} k x_1^2=\frac{1}{2} \times\left(5 \times 10^3\right) \times\left(5 \times 10^{-2}\right)^2$ $=6.25 \mathrm{~N} \mathrm{~m}$ $U_2=\frac{1}{2} k x_2^2=\frac{1}{2} \times 5 \times 10^3 \times(5+5)^2 \times 10^{-4}$ $=25 \mathrm{~N} \mathrm{~m}$ $\therefore$ Work done $=U_2-U_1=25.0-6.25$ $=18.75 \mathrm{~N} \mathrm{~m}$

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

Relevant Questions