Two springs with spring constants K_1 = 1500 | vedclass

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Question

(A) The potential energy $U$ stored in a spring stretched by a force $F$ is given by the formula $U = \frac{F^2}{2K}$,where $K$ is the spring constant

Vedclass · Accepted Answer

$2:1$

Vedclass · Answer

(A) The potential energy $U$ stored in a spring stretched by a force $F$ is given by the formula $U = \frac{F^2}{2K}$,where $K$ is the spring constant.Since the force $F$ is the same for both springs,we have $U \propto \frac{1}{K}$.Therefore,the ratio of potential energy stored in the two springs is $\frac{U_1}{U_2} = \frac{K_2}{K_1}$.Substituting the given values $K_1 = 1500 \, N/m$ and $K_2 = 3000 \, N/m$,we get $\frac{U_1}{U_2} = \frac{3000}{1500} = 2$.Thus,the ratio is $2:1$.

Two springs with spring constants $K_1 = 1500 \, N/m$ and $K_2 = 3000 \, N/m$ are stretched by the same force. The ratio of potential energy stored in the springs will be

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