# Conservation of Energy (Kinetic and Potential Energies)

1. Dec 10, 2012

### Pappers08

1. The problem statement, all variables and given/known data
A 5.00-kg block is set into motion up an inclined plane with an initial speed of 8.00 m/s. The block comes to rest after traveling 3.00m along the plane, which is inclined at an angle of 30° to the horizontal. For this motion, find the change in the block's kinetic energy, the change in the block's potential energy, the frictional force exerted on it, and the coefficient of kinetic energy.

2. Relevant equations
W= F*d
PE= mgh
KE= 1/2m(v squared)
Fnet=ma

3. The attempt at a solution
I haven't attempted it yet, because I'm stuck on how to start it. I missed the day of school we started going over this.

2. Dec 10, 2012

### cosmic dust

Since the block comes to rest, it looses all of it's kinetic energy. So the change will be equal to the initial kinetic energy mv2/2 .
The change in potential energy is mgΔh , where Δh is the change of height.That would be Δh = 3m sinθ.
When you do the above calculations you will see that the change in kinetic energy is not the same as the changde in potential energy. It looks like mechanical energy is not conserved, but that's because there is friction. The difference between change in kinetic energy and change in potential energy will be the work of friction. Find this difference and, since W = F Δx, the friction's will force will be W/Δx , where Δx = 3m

3. Dec 10, 2012

### Pappers08

So according to this, the block's kinetic energy would be 160J...the block's potential energy would be 73.5J...the frictional force would be 86.5N?

4. Dec 10, 2012

### cosmic dust

86.5 J is the work of friction. The force will be 87.6J/3m = 28.8 N

5. Dec 11, 2012

### Pappers08

Okay thanks! That really helps!