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