Question

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. 4 U
• B. 8 U
• C. 16 U
• D. U/4

Answer 1

Answer: (C)

Let extension produced in a spring be $x$ initially.
In stretched condition spring will have potential energy
$U=\frac{1}{2} k x^{2}$
Where $k$ is spring constant or force constant,
$\therefore \frac{U_{1}}{U_{2}}=\frac{x_{1}^{2}}{x_{2}^{2}}$ ...(i)
Given, $U_{1}=U, x_{1}=2\, cm , x_{2}=8\, cm$
Putting these values in Eq. (i), we have
$\frac{U}{U_{2}}=\frac{(2)^{2}}{(8)^{2}}=\frac{4}{64}=\frac{1}{16} $
$\therefore U_{2}=16 \,U$