A spring of spring constant $200 \mathrm{Nm}^{-1}$ is initially stretched by 10 cm from the unstretched…
A spring of spring constant $200 \mathrm{Nm}^{-1}$ is initially stretched by 10 cm from the unstretched position. The work to be done to stretch the spring further by another 10 cm is
3 J
6 J
9 J
12 J
Solution
$\mathrm{k}=200 \mathrm{~N}_{\mathrm{m}}{ }^{-1}, \mathrm{x}_1=10 \mathrm{~cm}, \mathrm{x}_2=10+10=20 \mathrm{~cm}$
$\therefore$ Work done to stretch the spring,
$\mathrm{W}=\frac{1}{2} k\left(\mathrm{x}_2^2-\mathrm{x}_1^2\right)=\frac{1}{2} \times 200 \times\left(20^2-10^2\right) \times 10^{-4}=3 \mathrm{~J}$