Two spring A and B are idential except that A is harder than B `(k_(A)gtk_(B))` if these are stretched by the equal force. In which spring will more work be done ?

Explanation

`F=k.x "so"x=F/K`
For same F, `W_(A)=1/2K_(A)x^(2)=(1F^(2))/(2K_(A)) and W_(B)=F^(2)/(2K_(B))`
`therefore W_(A)/W_(B)=K_(B)/K_(A)`
`"as" K_(A)gtK_(B) "so" W_(A)ltW_(B)`