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The potential energy of a long spring when stretched by 2 cm is $U$. If the spring is stretched by 8 cm, potential energy stored in it will be:
A
$16U$
B
$2U$
C
$4U$
D
$8U$
Explanation
Spring PE is $\frac{1}{2}kx^2$, so 4 times the stretch gives $16U$.
Detailed Solution
Potential energy stored in the spring $=\frac{1}{2}kx^2$
Now $\frac{1}{2}k(2)^2=U$ and $\frac{1}{2}k(8)^2=U'$ (say)
$\Rightarrow U'=\frac{64}{4}U=16U$