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Two springs P and Q are alike in all respects except that their spring constants satisfy K_P > K_Q. They are extended first through the same extension in case (a), and then by applying the same force in case (b). If the work done by the springs is W_P and W_Q, then the correct relation in cases (a) and (b), respectively, is

1. W_P = W_Q; W_P = W_Q
2. W_P > W_Q; W_Q > W_P
3. W_P < W_Q; W_Q < W_P
4. W_P = W_Q; W_P > W_Q

Correct answer: W_P > W_Q; W_Q > W_P

Solution

For equal extension x, W = (1/2)k x^2, so larger k gives more work: W_P > W_Q. For equal force F, W = F^2/(2k), so larger k gives less work: W_Q > W_P. Thus W_P > W_Q in case (a) and W_Q > W_P in case (b).

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