Block between springs, friction and work

[pleasehelpme6]

[PLAIN]https://wug-s.physics.uiuc.edu/cgi/courses/shell/common/showme.pl?cc/DuPage/phys2111/summer/homework/Ch-07-Work/IE_block_springs/9.gif

Homework Statement

A block of mass m = 1.5 kg slides between two springs, of spring constant kleft = 30 N/m and kright = 57 N/m. The distance between the relaxed springs is d = 2.8 m. The left spring is initially compressed a maximum of dleft = 0.7 m, and the block is released from rest. The first time the block hits the right spring, it compresses it a distance dright = 0.4 m Find the coefficient of sliding friction (M) between the block and the surface.

Homework Equations

W = (1/2)kx^2
Friction = M*N

The Attempt at a Solution

I used the equation (( W = 1/2 * kx^2 )) for both springs to get a net force of Wleft - Wright = 2.79 Nm.

In my understanding, this change in W caused by the work done by friction over the 2.8 m interval, so i solved W = F*d to get F = 0.9964 N.
((2.79 = F * 2.8))

0.9964 would be the force of friction, so i used the equations F = M*N and N = mg to solve for the friction constant (M).

The answer I came up with was 0.0667, but this is not right.
I also tried -0.0667, but this was also wrong.

Please help.

 

[pleasehelpme6]

In my understanding, this change in W caused by the work done by friction over the 2.8 m interval, so i solved W = F*d to get F = 0.9964 N.
((2.79 = F * 2.8))

I realized my mistake. Here, the distance is not 2.8, but is 2.8 + the compression distances of the springs ((2.8 + 0.4 + 0.7))

The equation should be...

2.79 = F * 3.9.

The answer is then 0.0486.

 

[PhanthomJay]

The friction force acts over a 2.8 +.7 + 0.4 = 3.9 m interval.
edit: You got it.