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(A) The potential energy stored in a spring stretched by distance $s$ is given by $U_1 = \frac{1}{2}ks^2 = 10 \ J$. When the spring is stretched by an

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$30$

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(A) The potential energy stored in a spring stretched by distance $s$ is given by $U_1 = \frac{1}{2}ks^2 = 10 \ J$. When the spring is stretched by an additional distance $s$,the total extension becomes $x = s + s = 2s$. The potential energy at this new extension is $U_2 = \frac{1}{2}k(2s)^2 = \frac{1}{2}k(4s^2) = 4 	imes (\frac{1}{2}ks^2)$. Substituting $U_1 = 10 \ J$,we get $U_2 = 4 	imes 10 \ J = 40 \ J$. The work done to stretch the spring from $s$ to $2s$ is the change in potential energy: $W = U_2 - U_1 = 40 \ J - 10 \ J = 30 \ J$.

$A$ spring is stretched by a distance $s$,and the potential energy stored in it is $10 \ J$. What is the work done to stretch the spring by an additional distance $s$ (in $J$)?

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