Two springs $A$ and $B$ are identical but $A$ is harder than $B(k_{A} k_{B})$. Let $W_{A}$ and $W_{B}$ represent the work done when the springs are stretched through the same distance and ${W'}_{A}$ and ${W'}_{B}$ are the work done when these are stretched by equal forces, then which of the following is true

Question

Two springs $A$ and $B$ are identical but $A$ is harder than $B(k_{A} > k_{B})$. Let $W_{A}$ and $W_{B}$ represent the work done when the springs are stretched through the same distance and ${W'}_{A}$ and ${W'}_{B}$ are the work done when these are stretched by equal forces, then which of the following is true

Dr. Abubacker · Accepted Answer

(b) $k_{A} > k_{B},\ x$ is the same $ herefore\frac{1}{2}k_{A}x^{2} > \frac{1}{2}k_{B}x^{2} \Rightarrow W_{A} > W_{B}$ Forces are the same $k_{A}x_{A} = k_{B}x_{B},$ As $k_{A} > k_{B},\ x_{A} < x_{B}$ $W_{A}^{'} = \frac{1}{2}\left( k_{A}x_{A} ight)x_{A}$ and $W_{B}^{'} = \frac{1}{2}\left( k_{B}x_{B} ight)x_{B}$ $ herefore W_{A}^{'} < W_{B}^{'}$; $ herefore W_{A} > W_{B}$ but $W_{A}^{'} < W_{B}^{'}$

Explanation

Reviewed by Dr. Abubacker

Want more practice?