Question

When a 4 kg mass is hung vertically on a light spring that obeys Hooke’s law, the spring stretches by 2 cms. The work required to be done by an external agent in stretching this spring by 5 cms will be (g = 9.8 m/sec²)(a) 4.900 joule(b) 2.450 joule(c) 0.495 joule(d) 0.245 joule

Answer 1

F=k*x...i)

W=1/2*k*x^2...ii)

From i- .K=1960 (N/m)

ii- W=2.45 (j)