Two similar springs $P$ and $Q$ have spring constants $k_P$ and $k_Q$, such that $k_P k_Q$. They are stretch

Question

Two similar springs $P$ and $Q$ have spring constants $k_P$ and $k_Q$, such that $k_P > k_Q$. They are stretched, first by the same amount (case a), then by the same force (case b). The work done by the springs $W_P$ and $W_Q$ are related as, in case (a) and case (b), respectively:

Accepted Answer

The work done ($W$) in stretching a spring is stored as elastic potential energy, given by the formula $W = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement . 1. **Case (a): Same displacement ($x_P = x_Q = x$)** Using $W = \frac{1}{2}kx^2$, since $x$ is constant, $W \propto k$. Given $k_P > k_Q$, it follows that **$W_P > W_Q$**. 2. **Case (b): Same force ($F_P = F_Q = F$)** We know $F = kx$, so $x = F/k$ . Substituting this into the work formula: $W = \frac{1}{2}k(F/k)^2 = \frac{F^2}{2k}$. Here, since $F$ is constant, $W \propto 1/k$. Given $k_P > k_Q$, it follows that **$W_P < W_Q$**. Thus, the correct relationships are $W_P > W_Q$ for case (a) and $W_P < W_Q$ for case (b).

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