Consider Example 6.8 taking the coefficient of friction , `mu` to be and calculate the maximum compression of the spring .

Explanation

In the presence of friction , both the spring force and the frictional force act so as to oppose the compression of the spring as shown in Fig .

From Work energy theorem ,
`DeltaK =W`
` :. K_(f) -K_(i) =W `
` 0-1/2 mv^(2) =W " "...(1)`
Wor done by resultant force,
`W = -1/2 kx_(m)^(2) -mumgs_(m)" "....(2)`
From equation (1) and (2) ,
` -1/2 mv^(2)=-1/2 kx_(m)6(2) -mumgx_(m)`
` :. 1/2 mv^(2) =1/2 kx_(m)^(2) +mumgx_(m)`
`mv^(2) =kx_(m)^(2)+2mumgx_(m)`
`10^(3)xx25 =6.25 xx10^(3)x_(m)^(2)+ 2xx0.5 xx10^(3) xx10 xx x_(m) [ :. g = 10 "ms"^(-2)] `
`10^(3) xx25 = 6.15 xx10^(3)x_(m)^(2)+2xx0.5xx10^(3)xx10xxx_(m) [ :. g = 10 ms^(-2)] `
`25 = 6.25 x_(m)^(2) +10x_(m)`
` :. 6.25 x_(m)^(2) +10x_(m)-25 = 0 ` ,
`a = 6.25 , b = 10 , c = -25`
`Delta =b^(2)- 4ac`
` = 100 +4 xx 6.25 xx25`
` = 100 + 625`
` = 725`
`x_(m) = (-b+sqrt(Delta))/(2a)`
` =(-10 pm sqrt(725))/(2xx6.25)`
`= (-10 pm 26.9)/(12.5)`
` :. x_(m) =(15.9)/(12.5) or x_(m) =-(36.9)/(12.5) ` is impossible
` :. x_(m) = 1.352 m `
` :. x_(m) approx 1.35 m `
Note :
If the two forces on the body consist of a conservative force `F_(c )` and a non -conservative force `F_(nc)` ,then from work ,
`(F_(c) +F_(nc)) Delta x = Deltak `
but `F_(c) Deltax = - DeltaV `
` :. Delta E = F_(nc )Deltax`
Where E is the total energy .For whole path .
`E_(f)-E_(i)=W_(nc)`
Where `W_(nc)` is the total work done by the non conservative forces over the path .