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Question

One end of an elastic spring of natural length L and spring constant k is fixed to a wall and the other end is attached to a block of mass m lying on a horizontal frictionless table. The block is moved to a position A, such that the spring is compressed to half its natural length and then released. What is the velocity of the block when it reaches position B, which is at a distance 3L4 from the wall.

A
L4km
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B
L22km
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C
L43km
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D
L2km
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Solution

The correct option is C L4√3kmAt position A, the compression of the spring is L2. At position B, the compression is L−3L4=L4. Therefore, loss of potential energy as the block moves from A to B is given by 12k(L2)2−12k(L4)2=3kL232 From conservation of energy, loss in P.E. = gain in K.E., i.e. 3kL232=12mv2 ⇒v=L4√3km

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