Shooting a Block up an Incline

[Winegar12]

Homework Statement

A block of mass m is placed in a smooth-bored spring gun at the bottom of the incline so that it compresses the spring by an amount x_c . The spring has spring constant k. The incline makes an angle theta with the horizontal and the coefficient of kinetic friction between the block and the incline is mu. The block is released, exits the muzzle of the gun, and slides up an incline a total distance L . Find L, the distance traveled along the incline by the block after it exits the gun. Ignore friction when the block is inside the gun. Also, assume that the uncompressed spring is just at the top of the gun (i.e., the block moves a distance x_c while inside of the gun). Use for the magnitude of acceleration due to gravity.
Express the distance L in terms of x_c,m ,g ,mu ,k , and theta.

Homework Equations

Work Energy Therom: E_final-E_initial=W_ext
E_initial=(1/2)k(x_c)²
W_friction=-Lmgcos(theta)mu
E_final=mg(L+x_c)sin(theta)

The Attempt at a Solution

I put these equations into the work energy therom and tried to solve for L, but I think I'm doing my algebra wrong, any suggestions? Here is what I have tried
L=[(1/2)kx²]/[mg(sin(theta)+cos(theta)mu]
L=[[(1/2)kx²]/[mg(sin(theta)+cos(theta)mu]]-x
L=[(kx²)-2mgxsin(theta)]/[2mg(mu)(cos(theta)+sin(theta))]
L=[[(1/2)kx²/[(mu)mgcos(theta)]]-x

I have tried a lot of different ways that are all wrong, can you see a place that I am doing my algebra wrong or any suggestions? Thanks!

 

[anathema]

Thank you for your post. Your approach using the work energy theorem is a good start, but there are a few things that need to be corrected.

Firstly, the work done by friction should be positive, as it acts in the opposite direction of motion. So the correct equation should be:

Wfriction = Lmgcos(theta)mu

Secondly, in your calculation for the final energy, you have used the wrong expression. It should be:

Efinal = (1/2)mv^2 + mg(L+xc)sin(theta)

where v is the velocity of the block at the end of the incline.

Finally, when setting the initial and final energies equal, be sure to include the work done by the spring, which is:

Wspring = (1/2)kxc^2

Putting all of these together, we get:

(1/2)mv^2 + mg(L+xc)sin(theta) = (1/2)kxc^2 + Lmgcos(theta)mu

Solving for L, we get:

L = [kxc^2 - 2mgxcsin(theta)] / [2mg(mu)(cos(theta)+sin(theta))] - xc

I hope this helps. Let me know if you have any further questions.