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Q.

When a 4 kg mass is hung vertically on a light spring that obeys Hooke's law, the spring stretches by 2 cm. The work required to be done by an external agent in stretching this spring by 5 cm will be (g =10  m/sec2)

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a

4.900 J

b

2.5 J

c

0.495 J

d

0.245 J

answer is B.

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Detailed Solution

K=Fx=402×10−2=2 ×103N/m Work done =12Kx2=12×2000×(0.05)2=2.5J
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