Consider example 25, taking the coefficient of friction `mu` to be 0.4 and calculate the maximum compression of the spring `(g=10ms^(-2))`

Explanation

Both the frictional force and the speing force act so as to oppose the compressin of the spring as shown in figure.

Use the work-energy theorem,
The change in `KE=Delta K=K_(t)-K_(i)=0-1/2mv^(2)`
The work done by the net force is `W=1/2kx_(m)^(2)mu mgx_(m)`
Equating the two, we get
`1/2mv^(2)=1/2Kx_(m)^(2)+mumgx_(m)`
`mumg=0.4xx1500xx10=6000N`
`Kx_(m)^(2)+2mumgx_(m)-mv^(2)=0" "("After rearranging the given equlatin")`
`x_(m)=(-2mumg+sqrt(4mu^(2)m^(2)g^(2)-4(k)(-mv^(2))))/(2K)("Taking + ve sign with square root as"x_(m)"is +ve")`
`=(-mumg+sqrt(mu^(2)m^(2)g^(2)+mkv^(2)))/(K)`
`=(-0.4xx1500xx10+sqrt((0.4xx1500xx10)^(2)+1500+7.5xx10^(3)xx10^(2)))/(7.5xx10^(3))=3.74m`
Which as expected, is less than the result in Example 25. If the two forces on the body consist of a conservation for `F_(c)` and a non-nocervative force `F_(nc,)` the conservation of mechanical energy formula will have to be modified by the W-E theorem,
`(F_(c)+F_(nc))Deltax=DeltaK`
But `F_(c)Deltax=-DeltaV`
Hence `Delta(K+V)=F_(nc)Deltax`
`DeltaE=F_(nc) Delta x`
Where E is the total mechanical energy. Over the path this assumes the form ` E_(f)-E_(i)=W_(nc)`
Where `W_(nc)` is the total work done by the non-conservative forces over the path. Unlike conservative force, `W_(nc)` depends on the path taken.