Two springs have their force constants `K_(1)` and `K_(2)` and they are stretched to the same extension. If `K_(2) gt K_(1)` work done is

Options

A. `W_(1)=W_(2)`
B. `W_(1) lt W_(2)`
C. `W_(1) gt W_(2)`
D. `W_(1)= W_(2)=0`

Answer

B

Explanation

`because` Spring constant `K=(F)/(x)`
`therefore K_(1)=(F)/(x_(1))=(F)/(K_(1)) and x_(2)=(F)/(K_(2))`
`therefore W_(1)=(1)/(2)K_(1)x_(1)^(2)=(1)/(2)K_(1)((F^(2))/(K_(1)^(2)))=(1)/(2)(F^(2))/(K_(1))`
and `W_(2)=(1)/(2)K_(2)x_(2)^(2)=(1)/(2)K_(2)((F^(2))/(K_(2)^(2)))=(1)/(2)(F^(2))/(K_(2))`
"sin"ce the same force is applied
`therefore (W_(1))/(W_(2))=(K_(2))/(K_(1)) but K_(1) gt K_(2)`
`therefore W_(2) gt W_(1) or W_(1) lt W_(2)`