# Spring problem

1. Oct 1, 2004

### 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 dont know the initial velocity.

How do I use X_0*K to calculate velocity?

2. Oct 1, 2004

### 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.

3. Oct 1, 2004

### merlinMan

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

4. Oct 1, 2004

### arildno

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:
$$\mu=\frac{K(x_{0}-\frac{L}{2})}{Mg}$$

Last edited: Oct 1, 2004