# Physics help using Work-Energy theorem

1. Mar 25, 2008

### ochatesme

1. The problem statement, all variables and given/known data
You are a member of an alpine rescue team and must get a box of supplies, with mass 2.60 kg, up an incline of constant slope angle 30.0$$\circ$$ so that it reaches a stranded skier who is a vertical distance 3.10 m above the bottom of the incline. There is some friction present; the kinetic coefficient of friction is 6.00×10$$^{-2}$$. Since you can't walk up the incline, you give the box a push that gives it an initial velocity; then the box slides up the incline, slowing down under the forces of friction and gravity. Take acceleration due to gravity to be 9.81 m/s$$^{2}$$.

Use the work-energy theorem to calculate the minimum speed (v) that you must give the box at the bottom of the incline so that it will reach the skier.

2. Relevant equations

Given that the problem revolves around the work-energy theorem I have set up an equation which I pulled from problems that were very similar to this one out of the textbook. I ended up with Ki + Ui + Wother = Kf + Uf, where K is kinetic energy, U is the gravitational potential energy and Wother is the force of friction in the system. The "i" in the equation stands for the initial position of the box at the bottom of the slope and the "f" in the equation is for the final position when the box has reached the top of the slope after overcoming friction.

3. The attempt at a solution

I made several attempts at this solution and I'll try my best to explain exactly what it was that I did, maybe you guys can help point out some of the errors I made because I am sure they are there. Given that Ui is at the bottom of the slope and there is no displacement in the vertical direction, I made that part of the system "0". I also made Kf "0" due to the fact that at the final position of the box at the top of the slope, the box no longer has kinetic energy. Thus far my equation looked something like this "Ki + 0 + Wother = 0 + Uf". After this, I calculated Uf to be 39.5343 J. One of the steps that I think I'm really messing up on is calculating the amount of work done by the force of friction, what I'm doing on this step is taking the coefficient of friction and multiplying it by the normal force, which ends up looking like (6.00 x 10^-2)(2.60kg)(9.81m/s^2)(sin30) which calculates out to 0.76518 J. When I plug all this back into the original equation to try to solve for V I end up with the wrong answer each and every single time. I'm 100% sure that the mistake I'm making comes when I'm making an attempt to calculate the work required when going through the force of friction. Thanks in advance for your help!

Last edited: Mar 25, 2008
2. Mar 25, 2008

### ochatesme

Someone..........anyone? If someone could lead me in the right direction in figuring out how to calculate the work done by the force of friction (W other) I would really appreciate it!

3. Mar 26, 2008

### physixguru

First analyze the forces acting on the box of supplies as it moves up the incline.
Show me your free body diagram.

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