The work done by an external agent in pulling a spring from a deformation of `10cm` to `15cm` is `W_1`. When pulled, the spring from a deformation of `15cm` to `20cm`, the work done is `W_2`. Find `W_1//W_2`.

Explanation

`W_1=1/2k(x_2^2-x_1^2)`
where `x_2=15` and `x_1=10cm`
`=1/2k[(15)^2-(10)^2]=125/2k`
Similarly, `W_2=1/2k[(20)^2-(15)^2]=(175k)/(2)` (ii)
From Eqs. (i) and (ii), `W_2/W_1=((175//2)k)/((125//2)k)=7/5`