Height of Hill with Potential and Kinetic Energy

[Dirst]

Homework Statement

A child and sled with a combined mass of 50.0 kg slide down a frictionless hill. If the sled starts from rest and has a speed of 12.0 m/s at the bottom, what is the height of the hill?

Homework Equations

KE = 1/2 m v^2
PE = mgh

The Attempt at a Solution

KE = 1/2 (50.0) (12.0^2)
KE = 3600 N

No idea what to do. :(

 

[jegues]

How much potential/kinetic energy do you have at the top of the hill?

How much potential/kinetic energy do you have at the bottom of the hill?

Remember, energy is conserved.

Kt + Pt = Kb + Pb

 

[Dirst]

Well at the top:

KE = 1/2*(50.0)*(0)
KE = 0 N

PE = (50.0)*(9.8)*H <-- Need to find this... So I guess I can't do PE.

Bottom:

KE = 1/2*(50.0)*(12.0^2)
KE = 3600 N

Again, you can't do PE because you don't know H...

 

[jegues]

You were able to find the correct kinetic energy at the bottom and top of the hill. (It's in Joules, not Newtons). Energy is conserved throughout the intial and final states.

mgh + 0 = 3600J + 0,

Solve h.

 

[Dirst]

According to that,

H = 3600 J / ((9.8 m/s^2)*(50.0 kg))
H = 7.35 m

Is Pb = 0 because the height would equal 0 m?

 

[denverdoc]

This is right, it is just defined as a zero point for calculation purposes. The real zero would be at the Earth's center of course.

 

[Dirst]

Thank you both. On to my next problem... :(