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Question

# A block is attached to a spring having equilibrium position at x=0. The block oscillates between x=−4 m and x=4 m. Find the change in potential energy stored in the spring when the block moves from x=2 to x=−3. Given the spring constant is 5 N/m.

A
12.5 J
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B
32.5 J
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C
37.5 J
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D
10 J
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Solution

## The correct option is A 12.5 JEnergy stored when spring is at x=2 m is U1=12kx2=12×5×22 U1=10 J Energy stored when spring is at x=−3 m is U2=12kx2=12×5×(−3)2 U2=22.5 J Change in potential energy of the spring =U2−U1 =22.5−10=12.5 J

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