# Shooting a block up an incline - work energy

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 g for the magnitude of acceleration due to gravity.
Express the distance L in terms of mu, theta, g, x_c, m, and k.

My energy equation is as follows:

0.5k(x_c)^2 = 0.5mv^2 + (mu)mg((x_c) + L)cos(theta)
0.5k(x_c)^2 = (mu)mg((x_c) + L)cos(theta)
[0.5k(x_c)^2]/[(mu)mgcos(theta)] = x_c + L
L = [0.5k(x_c)^2]/[(mu)mgcos(theta)] - x_c

I know you have to add X_c to L to find the total distance the block is moved by the spring, and I really don't understand why this is wrong. Anyone know?

Last edited:

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It seems to me that you have included friction while the block is in the gun, if I am reading that correctly.
But anyway, this travels up an incline, did you get all the energy associated with your system?

0.5k(x_c)^2 = mg((x_c) + L)sin(theta) + (mu)mgLcos(theta)

I guess I forgot the y component

But that leaves me with:

L=[k(x_c)^2 - 2mg(x_c)sin(theta)]/[2mg[(mu)cos(theta) + sin(theta)]]

Which seems to me way too complicated an expression.

0.5k(x_c)^2 = mg((x_c) + L)sin(theta) + (mu)mgLcos(theta)

I guess I forgot the y component

But that leaves me with:

L=[k(x_c)^2 - 2mg(x_c)sin(theta)]/[2mg[(mu)cos(theta) + sin(theta)]]

Which seems to me way too complicated an expression.
lol, that is not too bad of an expression!
*edit* - the first equation looks okay to me, I didnt check your algebra in the next step.