# Spring on an incline

1. Oct 10, 2006

### PascalPanther

The situation is as follows:

There is friction, but no friction while in the barrel of the spring gun.

I am suppose to derive an equation to solve for L, using x(c), m, g, theta, mu, and k.

What I did so far:

I did a FBD, I see the following forces:
1. Force of the spring from the spring gun, 2. force of the weight parallel to the plane, 3. force of kinetic friction

U(spring) = 1/2 k*x(c)^2
F(weight) = m*g*sin(theta)
F(friction) = kinetic friction constant* normal force
= mu(k) * m*g*cos(theta)

W=F*s = 1/2 * m*v^2 - 0
F(net) = U - F(w) - F(f)

s = F/W
= [U - F(w) - F(f)] / [(1/2) * m * v^2]
= [(1/2)k*x(c)^2 - (mgsin(theta)) - mu*mgcos(theta))]/ [(1/2)/m*v^2]

Since I can't use v in the equation;
U(spring) = KE = (1/2)k*x^2 = (1/2)m*v^2
v = Sqrt((k*x^2)/m)

L = s
L = [(1/2)k*x(c)^2 - (mgsin(theta)) - mu*mgcos(theta))]/ [(1/2)/m*((k*x^2)/m)]

What did I do wrong here? I think it's more than one thing wrong.

Last edited: Oct 10, 2006
2. Oct 10, 2006

### OlderDan

I can't quite follow what you are doing, but I see some dimensional inconsistencies. For example
F(net) = U - F(w) - F(f)
cannot be. U is an energy and the Fs are forces. These do not add.

You seem to recognize that there is a force in the problem that is going to do work that will take mechanical energy from the mass. That is good. Focus on the mechanical energy. What would happen if there were no friction? What changes when the friction is present?

3. Oct 10, 2006

### PascalPanther

I think I need to start over.

So there should be two different parts to this problem.
First part, is just inside the spring gun barrel:

U(spring) = (1/2)*k*x^2
U(grav) = sin(theta)mg(-x)

KE = U(spring) + U(grav)
(1/2)*m*v^2 = (1/2)*k*x^2 - sin(theta)mgx

v^2= [(k*x^2)/m] - [2*sin(theta)*g*x]
v(1) = Sqrt([(k*x^2)/m] - [2*sin(theta)*g*x])
Which is the velocity as it exits the gun barrel...

After this it is no longer a spring problem.
If there was no friction, I think it will be just:
K(1) + U(1) = K(2) + U(2)

But there is friction so:
K(1) + U(1) + W(other) = K(2) + U(2)
(1/2)*m*v^2 + 0 + W(other) = 0 + U(2)
(1/2)*m*v^2 + W(other) = m*g*h

W(other) = F(friction)*s
F(friction) = mu*N = mu*(m*g*cos(theta))

I'm a bit confused at this part, I think s = L. But I don't know "h" so I can't solve for W(other) to then use to find s... Am I still going about this wrong?

4. Oct 10, 2006

### OlderDan

You are close. There is a simple relationship between your h and L. Look back at what you did when the block was in the gun.

5. Oct 10, 2006

### PascalPanther

h is equal to (L+x(c))*sin(theta)? Or is there something else? I don't think I could solve for L if that was the case...

6. Oct 10, 2006

### OlderDan

It looks to me that your h is the increase in height relative to the end of the gun, so it would be just
h = L*sin(theta)

7. Oct 10, 2006

### PascalPanther

(1/2)*m*v^2 + W(other) = m*g*h

(1/2)*m*v^2 + mu*(m*g*cos(theta))*L = m*g*L*sin(theta)

((1/2)*v^2)/L = -mu*(g*cos(theta)) + g*sin(theta)

L/((1/2)*v^2) = 1/(-mu*(g*cos(theta))) + 1/(g*sin(theta))

L = ((1/2)*v^2)/(-mu*(g*cos(theta))) + ((1/2)*v^2)/(g*sin(theta))

Plugging in how I got v... should get me how to find L?

------
I ended up also looking at the hints for this problem, and it gave me after solving the parts:

E(final) - E(initial) = W(friction)

E(final) = m*g*(L + x)*sin(theta)
E(initial) = (1/2)k*x^2
W(friction) = -k.coeff*(m*g*cos(theta))*L

m*g*(L+x)*sin(theta) - (1/2)k*x^2 = -k.coeff*(m*g*cos(theta))*L

Does my other way sound compatible to this way? Is there even a way to solve for L in the hinted way?

8. Oct 10, 2006

### OlderDan

Your way was OK, but it was not the most direct way. There was no need to find the velocity at the end of the gun. All that matters is that the differeence between the final gravitaional potential energy and the intial spring potential energy equals the work done against the system by friction. Since you only care about the energies at the beginning and at the end, where the block is not moving, the intermediate velocities are of no concern.

9. Oct 10, 2006

### PascalPanther

Since my first way produces a huge equation, how would I go about moving the L to one side for the second equation? I think I must be missing some algebra trick. I cannot find a way to get (L-x) and L to work.

10. Oct 10, 2006

### OlderDan

Expand the term with L+x into two terms using the distributive property. Then group your L terms on one side of the equation and solve for L.