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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 the potential energy stored in it is :-
A
$4U$
B
$8U$
C
$16U$
D
$\dfrac{U}{4}$
Detailed Solution
Potential energy stored in a spring stretched by $x$: $U = \dfrac{1}{2}Kx^2$
$U = \dfrac{1}{2}K(2)^2$ and $U' = \dfrac{1}{2}K(8)^2$
$\Rightarrow \dfrac{U'}{U} = \left(\dfrac{8}{2}\right)^2 = 16$
$\Rightarrow U' = 16U$
$U = \dfrac{1}{2}K(2)^2$ and $U' = \dfrac{1}{2}K(8)^2$
$\Rightarrow \dfrac{U'}{U} = \left(\dfrac{8}{2}\right)^2 = 16$
$\Rightarrow U' = 16U$