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(C) The potential energy $U$ of a spring stretched by a distance $x$ is given by the formula: $U = \frac{1}{2} k x^2$.Since the spring constant $k$ is

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$16$

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(C) The potential energy $U$ of a spring stretched by a distance $x$ is given by the formula: $U = \frac{1}{2} k x^2$.Since the spring constant $k$ is constant,the potential energy is directly proportional to the square of the displacement: $U \propto x^2$.Initially,for $x_1 = 1 \, cm$,the potential energy is $U_1 = U$.When the spring is stretched by $x_2 = 4 \, cm$,the new potential energy $U_2$ is:$U_2 = \frac{1}{2} k (x_2)^2 = \frac{1}{2} k (4 \, cm)^2 = 16 	imes (\frac{1}{2} k (1 \, cm)^2) = 16 U$.Therefore,the potential energy becomes $16U$.

$A$ 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 (in $U$)

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