Work done for a certain spring when stretched through 1 mm is 10 joule. The amount of work that must be done on the spring to stretch it further by 1 mm is

Options

A. 30J
B. 40J
C. 10J
D. 20J.

Answer

A

Explanation

Work done for a certain spring = Stored potential energy in the spring
Initial potential energy of spring, `U_(f) = 10 J`
`U_(i) = (1)/(2) k x_(i)^(2)`
Required amount of work done on the spring = Change in P.E. of the spring
`=U_(f) - U_(i) = (1)/(2) k x_(f)^(2) - (1)/(2) x_(i)^(2)`
`=(1)/(2) k (2 x_(i) )^(2) - (1)/(2) k x_(i)^(2) = 3 xx (1)/(2) k x_(i)^(2) " "(because x_(f) = 2x_(i))`
`=3 U_(i) = 3 xx 10 = 30` J.