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 = 9.8 metre/sec2)

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4.900 joule

2.450 joule

0.495 joule

0.245 joule

answer is B.

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Detailed Solution

K = Fx = 402×10-2 = 0.2 N/mWork done = 12Kx2 = 12×(0.2)×(0.05)2 = 2.5 J

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