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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