Two springs have their force constant as `k_(1)` and `k_(2) (k_(1) gt k_(2))`. When they are streched by the same force.

Options

A. more work is done on B i.e., `W_(B) gt W_(A)`
B. more work is done on B i.e., `W_(A) gt W_(B)`
C. work done on A and B are equal
D. work done dependsupon the way in which they are streched.

Explanation

`F = k_(A) x_(A) = k_(B) x_(B)`
`W_(A) = (1)/(2) k_(A) x_(A)^(2), W_(B) = (1)/(2) k_(B) x_(B)^(2)`
`(W_(A))/(W_(B)) = (k_(A))/(k_(B)) (x_(A)^(2))/(x_(B)^(2)) = (k_(A))/(k_(B)) (k_(B)^(2))/(k_(A)^(2)) = (k_(B))/(k_(A))`
Hence, `W_(B) gt W_(A)`. So more work is done on `B` .
`ME_(max) = 1//2 m (2u)^(2) = 2 mu^(2)`.