Consider Example 8, taking the coefficient of friction, `mu`, to be 0.5 and calculate the maximm compression of the spring.

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.