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

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0.1 joule

0.2 joule

0.3 joule

0.5 joule

answer is A.

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Detailed Solution

Δ P.E. =12kx22−x12=12×10(0.25)2−(0.20)2=5×0.45×0.05=0.1J

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