Two springs of spring constants 1500 N/m and | vedclass

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Question

(C) The potential energy $U$ stored in a spring is given by the formula $U = \frac{F^2}{2k}$,where $F$ is the applied force and $k$ is the spring cons

Vedclass · Accepted Answer

$2:1$

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(C) The potential energy $U$ stored in a spring is given by the formula $U = \frac{F^2}{2k}$,where $F$ is the applied force and $k$ is the spring constant.Since both springs are stretched with the same force $F$,the potential energy is inversely proportional to the spring constant: $U \propto \frac{1}{k}$.Therefore,the ratio of the potential energies is $\frac{U_1}{U_2} = \frac{k_2}{k_1}$.Given $k_1 = 1500 \ N/m$ and $k_2 = 3000 \ N/m$,we have:$\frac{U_1}{U_2} = \frac{3000}{1500} = \frac{2}{1}$.Thus,the ratio is $2:1$.

Two springs of spring constants $1500 \ N/m$ and $3000 \ N/m$ respectively are stretched with the same force. They will have potential energy in the ratio

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