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Question

# A massless spring with a force constant K=40N/m hangs vertically from the ceiling. A 0.2kg block is attached to the free end of the spring and held in such a position that the spring has its natural length and suddenly released. The maximum elastic strain energy stored in the spring is (take g=10 m/s2):

A
0.1J
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B
0.2J
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C
0.05J
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D
0.4J
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Solution

## The correct option is A 0.2JThe decrease in potential energy of 0.2kg mass will be gain in elastic energy of spring⇒(P.E)spring=12kx2...(i)Decrease in potential energy of 0.2kg mass =0.2gh...(ii)From equation (i) and (ii):12kx2=.2ghx=2mgkDecrease in potential energy, mgx=mg×(2mgk) =2(mg)2k =2×(0.2)2×(10)240 =420=0.2J

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