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
-Kx-\mu{Mg}=Ma
Here, "x" is the compressed length of the spring, and
x(t=0)=-x_{0}
and
x(t=t_{L})=L-x_{0}
(t_{L} is the time when the system stops; when a distance L has been traversed)
-\mu{Mg} 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 t=t_{L}:
a) -Kxv-\mu{Mg}v=Mav
b) -\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})
Or, by recognizing:
v(t=0)=v(t=t_{L})=0
we gain by rearranging:
\mu{MgL}=\frac{K}{2}(x_{0}^{2}-(L-x_{0})^{2})=\frac{KL}{2}(2x_{0}-L)
Or:
\mu=\frac{K(x_{0}-\frac{L}{2})}{Mg}