Two springs of force constant `100N//m` and `150N//m` are in series as shown the block is pulled by a distance of `2.5` cm to the right from equilibrium position what is the ratio of work done by the spring at left to the work done by the spring at right:

Options

A. `3/2`
B. `2/3`
C. `0.2`
D. none of these

Answer

A

Explanation

`k_(eq)=(100xx150)/250=60N//m`
`f=k_(eq) x=60xx2.5/100=3/2N`
for left spring `x_(1)=3/(2(100))`
for right spring `x_(2) =3/(2(150))`
`So (1/2(100)(3/2)^(2) (1/100)^(2))/(1/2(150)(3/2)^(2)(1/150)^(2))=150/100=3/2`