Two similar springs `P` and `Q` have spring constant `K_(P)` and `K_(Q)` such that `K_(P) gt K_(Q)`. They are stretched, first by the same amount (case a), then the same force (case b). The work done by the spring `W_(P)` and `W_(Q)` are related as, in case (b), respectively

Options

A. `W_(P) gt W_(Q), W_(Q) gt W_(P)`
B. `W_(P) lt W_(Q), W_(Q) lt W_(P)`
C. `W_(P) = W_(Q), W_(P) gt W_(Q)`
D. `W_(P) = W_(Q), W_(P) = W_(Q)`

Answer

`(1)`

Explanation

Here, `K_(P) gt K_(Q)`
Case (a) : Elongation(x) in each spring is same.
`W_(P) = 1/2(K_(P) x^(2)), W_(Q) = 1/2 (K_(Q)x^(2))`
`therefore W_(P) gt W_(Q)`
Case (b) : Force of elongation is same
So, `x_(1) = F/(K_(p)) and x_(2) = (F)/(K_(Q))`
` W_(p) = 1/2 (K_(p)x_(1)^(2)) = 1/2(F^(2))/(K_(P))`
`W_(Q) = 1/2(K_(Q)x_(2)^(2)) = 1/2(F^(2))/(K_(Q))`
`therefore W_(P) lt W_(Q)`