When a 4 kg mass is hung vertically on a light string 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(metrs)/(sec^2)`)

Options

A. `4.900 joule`
B. `2.450 joule`
C. `0.495 joule`
D. `0.245 joule`

Answer

B

Explanation

`K=(F)/(x)=(40)/(2xx10^-2)=0.2(N)/(m)`
Work done `=(1)/(2)Kx^2=(1)/(2)xx(0.2)xx(0.5)^2=2.5J`