Question

A long spring when stretched by $xcm$ has a potential energy $v$. On increasing the stretching to $nxcm$, the potential energy stored in the spring will be

• A. $nv$
• B. $n^{2}v$
• C. $n{v}^3$
• D. $n{v}^{-2}$

Answer 1

Answer: (B)

$n^{2}v$

Potential energy$=\frac{1}{2}k{x}^2=v$
When the stretching is $nx$ cm,
PE$=\frac{1}{2}k(nx{)}^2=n^2[\frac{1}{2}k{x}^2]=n^{2}v$