Calculating Kinetic Friction Using F=ma and X_0*K

[merlinMan]

a block of mass M is held on a horizontal table against a compressed horizontal (massless) spring. When released from rest, the block is launched along the tabletop and eventually comes to a stop after sliding a total distance L. The initial spring compression is x_0 and the spring constant is K. Calculate teh coefficient of kinetic friction between the block and the ttable top in terms of the variables M, L, K, and X_0.

Im lost. I know F=ma. I know x_0*k is the force the spring is dishing out. I know that the frictional force is Mu*M*g. I tried to put it together with v^2 = v_0^2 + 2aD. The problem with that is I don't know the initial velocity.

How do I use X_0*K to calculate velocity?

 

[arildno]

Hint:
Use an equation relating the mechanical energy of the system at 2 different places and the work of friction done over that interval.

 

[merlinMan]

Could you walk me through that . . . I'm still lost.

 

[arildno]

OK, let's start with Newton's 2.law, and derive the energy equation.
1.Newton's 2.law
[tex]-Kx-\mu{Mg}=Ma[/tex]
Here, "x" is the compressed length of the spring, and
[tex]x(t=0)=-x_{0}[/tex]
and
[tex]x(t=t_{L})=L-x_{0}[/tex]
([tex]t_{L}[/tex] is the time when the system stops; when a distance L has been traversed)
[tex]-\mu{Mg}[/tex] is the frictional force, whereas a is the acceleration of the system.

2. Derivation of energy equation
We a) multiply the above equation with velocity v, and
b) integrate from t=0 to [tex]t=t_{L}[/tex]:
a) [tex]-Kxv-\mu{Mg}v=Mav[/tex]
b) [tex]-\frac{K}{2}x(t=t_{L})^{2}+\frac{K}{2}x(t=0)^{2}-\mu{MgL}=\frac{M}{2}(v(t=t_{L})^{2}-v(t=0)^{2})[/tex]
Or, by recognizing:
[tex]v(t=0)=v(t=t_{L})=0[/tex]
we gain by rearranging:
[tex]\mu{MgL}=\frac{K}{2}(x_{0}^{2}-(L-x_{0})^{2})=\frac{KL}{2}(2x_{0}-L)[/tex]
Or:
[tex]\mu=\frac{K(x_{0}-\frac{L}{2})}{Mg}[/tex]