A block of mass m is welded with a light spring of stiffness k. The spring is initially relaxed. When the wedge fitted moves with an acceleration a, as shown in figure. `8.60`, the block slides through a maximum distance `l` relative to the wedge. If the coefficient of kinetic friction between the block and the wedges is `mu`, find the maximum deformation `l` of the spirng by using work-energy theorem.

Explanation

In this case, we are given the block slides through a distance `l` relative to the wedge. We can apply work-energy theorem from the frame of reference of wedge.

While the block is displaced from point 1 to point 2, work done by all the forces are given as follows:
Work: `W_(gr)=-mglsin theta`, `W_(sp)=(-Kl^2)/(2)`
`W_f=(muN)*l`
`N=mg cos theta+ma sin theta`
`W_f=-mu mgl cos theta-mu ma sin theta`
and `W_(pseudo)=(ma cos theta)l= mal cos theta`
W-E theorem: Summing upall workds, the total done is
`W_(t otal)=W_(gr)+W_(sp)+W_f+W_(pseudo)`
`W=-(mgl sin theta)/(1)-(mu mgl cos theta)/(3)+(mal cos theta)/(4)-(1/2kl^2)/(2)+Fl-(mu mal sin theta)/(3)`
Since the block was at rest at 1 and will remain at rest relative to the wedge, we can write `DeltaK=0` (relative to the wedge). Using work-energy theorem relative to the wedge, we have `W=DeltaK=0`.
Substituting `W=0`, we have
`-mg//sintheta-1/2kl^2-mumgcostheta-mumalsintheta+malcostheta=0`
`l=2/k[ma(costheta-musintheta)-mg(sintheta+mucostheta)]`
`=(2m)/(k)[a(costheta-musintheta)-g(sintheta+mucostheta)]`