When a long spring is stretched by $2 \mathrm{~cm}$, its potential energy is $U$. If the spring is stretched…
When a long spring is stretched by $2 \mathrm{~cm}$, its potential energy is $U$. If the spring is stretched by $10 \mathrm{~cm}$, the potential energy stored it will be:
$\frac{U}{5}$
$5 U$
$10 U$
$25 U$
Solution
$U=\frac{1}{2} K(2)^2$
$U^{\prime}=\frac{1}{2} K(10)^2$
From above equations
$U=25 \mathrm{~U}$