5.Work, Energy, Power and Collision
easy

A spring with spring constant k when stretched through $1\, cm$, the potential energy is $U$. If it is stretched by $4 \,cm.$ The potential energy will be

A

$4U$

B

$8U$

C

$16 U$

D

$2U$

Solution

(c)Potential energy $U = \frac{1}{2}k{x^2}$
$U \propto {x^2}$[if k = constant]
If elongation made $4$ times then potential energy will become $16\, times$.

Standard 11
Physics

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