Question

A spring with spring constant $k$ when compresses by $1cm$, the potential energy stored is $U$ If it is further compressed by $3cm$ then change in its potential energy is

• A. $3U$
• B. $9U$
• C. $8U$
• D. $15U$

Answer 1

The correct option is D $15U$

We know that,

$U=\frac{1}{2}k{x}^2$

$\Rightarrow U=\frac{1}{2}k(1{)}^2=\frac{k}{2}$

Total compression $x^{\prime }=1+3=4$

Thus the new PE is,

$U^{\prime }=\frac{1}{2}k(x^{\prime }{)}^2=\frac{1}{2}k\times 4^2=\frac{16K}{2}=16U$

Change in potential energy $\Delta U=16U-U=15U$