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Q.
The potential energy of a spring increases by $15\, J$ when stretched by $3\, cm$. If it is stretched by $4\, cm$, the increase in potential energy is
Work, Energy and Power
Solution:
Potential energy of spring, $PE =\frac{1}{2} k x^{2}$
$\Rightarrow PE \propto x^{2}$
$\because \frac{( PE )_{2}}{( PE )_{1}}=\left(\frac{x_{2}}{x_{1}}\right)^{2}$
$\Rightarrow ( PE )_{2}=( PE )_{1} \times\left(\frac{x_{2}}{x_{1}}\right)^{2}$
$=15 \times\left(\frac{4}{3}\right)^{2}=27\, J$