Question

Two springs have force constants `k_(1) " and " k_(2) (k_(1) gt k_(2))`. On which spring is more work done, if (i) they are stretched by the same force and (ii) they are stretched by the same amount ?

Answer 1

(i) Let each spring be stretched through a distance y.
Work done on second spring `=(1)/(2) k_(2)y^(2)`
Work done on first spring `(1)/(2)k_(1)y^(2)`. But `k_(1) gt k_(2)` (given)
`therefore (1)/(2)k_(1)y^(2) gt (1)/(2)k_(2)y^(2)`
So, more work is done on the first spring.
(ii) Let each spring be stretched by the same force F. Let `y_(1)` and `y_(2)` be the extension in the first and second springs respectively.
Then, `y_(1)=(F)/(k_(1))`
and `y_(2)=(F)/(k_(2))`
Work done on first spring `=(1)/(2)k_(1)y_(1)^(2)=(1)/(2)k_(1)((F)/(k_(1)))^(2)=(1)/(2)(F^(2))/(k_(1))`
Similarly, the work done on the second spring `(1)/(2)(F^(2))/(k_(2))`
But `k_(1) gt k_(2) " "`(given)
`therefore (1)/(2)(F^(2))/(k_(1)) lt (1)/(2)(F^(2))/(k_(2))`
So, more work is done on the second spring.