One end (say B) of a massless spring having force constant k is attached to a block lying on a smooth surface while the other end A is pulled by an external force At some instant the velocities of end A and B of the spring are `v_(A)` and `v_(B)` respectively. If the energy of the spring is increasing at the rate of `P J//s` find the instantaneous tension and stretch in the spring.

Explanation

Let l be the natural length of the spring and `x_(A)` and `x_(B)` be the position of ends A and B at time t, then stretched length of the spring will be

`l' = x_(A)-x_(B)`
and so the stretch
`x=l'-l=(x_(A)-x_(B)-l)`
and so, `F=kx=k(x_(A)-x_(B)-l)`
and `U=1/2kx^(2)=1/2k(x_(A)-x_(B)-l)^(2)`....(ii)
Diferentiating eqn (ii) with respect to time,
`(dU)/(dt)=k(x_(A)-x_(B)-l)((dx_(A))/(dt)-(dx_(B))/(dt))` ...(iii)
But as `(dU)/(dt)=P,(dx_(A))/(dt)=v_(A)and (dx_(B))/(dt)=v_(B)`(given)
So, Eqn (iii) in the light of (i) reduces to
`P=F(v_(A)-v_(B))`
So, tension `F=P/((v_(A)-v_(B)))`
and Strectch `x=F/k=P/k(v_(A)-v_(B))`