A block of mass `m` is moving with a speed `v` on a horizontal rough surface and collides with a horizontal mounted spring of spring constant `k` as shown in the figure .The coefficient of friction between the block and the floor is `mu` The maximum compression of the spring is

Options

A. `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`
B. ` (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`
C. `- (mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) - mkv^(2))`
D. `(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`

Answer

A

Explanation

In pressent of friction both the spring force and the frictional act so as to oppose the compression of the spring
Work done by the net force
`W = -(1)/(2) kx_(m)^(2) - mu mhs_(m)`
where `x_(m)` is the maximum compression of the spring change in kinetic energy
`Delta K = K_(1) - K_(1) = 0 - (1)/(2) mv^(2)`
Accerding to worik energy therem
`W = Delta K`
`- (1)/(2) kx_(m)^(2) - mu mgx_(m) = -(1)/(2)mv^(2)`
` kx_(m)^(2) + mu mgx_(m) = (1)/(2)mv^(2)`
` kx_(m)^(2) + 2mu mgx_(m) = -mv^(2)= 0`
` x_(m)^(2) +(2 mu mgx_(m))/(k) - (mv^(2))/(k) = 0`
it is a quadratic equation in `x_(m)`
Solving this equation for `x_(m)` is positive we get
`x_(m) = -(mu mg)/(k) + (1)/(k) sqrt((mu mg)^(2) + mkv^(2))`