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Question

A spring of force constant 10 N/m has an initial stretch 0.20 m. In changing the stretch to 0.25 m, the increase in potential energy is about-

A
0.1 J
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B
0.2 J
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C
0.3 J
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D
0.5 J
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Solution

The correct option is A 0.1 J

Potential energy of object at initial stretch

P1 = 12k×2 = 12×10×0.2×0.2 = 0.2 J

Potential energy of object at changed stretch

P2 = 12k×2 = 12×10×0.25×0.25 = 0.3 J

difference in potential energy = P2 P1 = 0.3 J 0.2 J = 0.1 J


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