Consider example 14, 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 spring force act so to oppose the compression of the spring as shown in figure.

Use the work - energy theorem,
The change in `KE=DeltaK=K_(f)-K_(i)=0-(1)/(2)mv^(2)`
The work done by the net force is `W=-(1)/(2)Kx_(m)^(2)-mu mg x_(m)`
Equating the two, we get
`(1)/(2)mv^(2)=(1)/(2)Kx_(m)^(2)+mu mg x_(m)`
`mumg=0.4xx1500xx10=6000N`
`Kx_(m)^(2)+2mumg x_(m)-mv^(2)=0" (After rearranging the given equation)"`
`x_(m)=(-2mu mg+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)-4(k)(-mv^(2))))/(K)`
`=(-0.4xx1500xx10+sqrt((0.4xx1500xx10)^(2)+1500xx7.5xx10^(3)xx10^(2)))/(7.5xx10^(3))`
`=3.75m`
Which, as expected, is less than the result in Example 14. If the two forces on the body consist of a conservative force `F_(c)` and a non - conservative 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)Deltax`
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.