A spring stores 1J of energy for a compression of 1 mm. The additional work to be done to compress it further by 1 mm is

Options

A. 1J
B. 2J
C. 3J
D. 4J

Answer

C

Explanation

Given,
`W _(1) = 1J, x =1 1 xx 10 ^(-3) m`
`k = ?` (where k= spring constant )
`W _(1) = 1/2 kx ^(2)`
`1= (1)/(2) xx k xx (1 xx 1 ^(-3)) ^(2)`
`k = 2 xx 10 ^(6)`
`W_(2) = (1)/(2) kx _(1) ^(2) `
`= (1)/(2) xx 2 xx 10 ^(6) xx (1 xx 10 ^(-3) + 1 xx 10 ^(-3)) ^(2)`
`W _(2) = 4 xx 10 ^(-6) xx 10 ^(6) = 4 J`
The additional work done `= W _(2) - W_(1) = 4 (-1) = 3 J`