Car rolling down a hill. How high will it go up another incline?

[Bob Loblaw]

Homework Statement

A 684 kg automobile is moving at 13.2 m/s at a height of 5.0 m above the bottom of a hill when it runs out of gasoline. The car coasts down the hill and then continues coasting up the other side until it comes to rest. Ignoring frictional forces and air resistance, what is the value of h, the highest position the car reaches above the bottom of the hill?

http://www.webassign.net/grr/p6-31alt.gif

Homework Equations

Mechanical energy is conserved so Einitial=Efinal. Uinitial+Kinitial=Ufinal+Kfinal.
mgyinitial+1/2mvinitial^2=mgyfinal+1/2mvfinal^2

The Attempt at a Solution

I understand that the highest point will be reached when vfinal is equal to zero. However - I am having great difficulty manipulating the equations to allow for this equation. Any help or guidance will be appreciated!

 

[berkeman]

Another way to approach this problem is to remember that the total energy (KE + PE) is constant when there are no friction/etc losses... What is the total energy of the car initially? What will the KE be when it comes to rest at the top of the arc on the opposite side?

 

[Bob Loblaw]

So examining the problem your way I have:

KE=1/2(684kg)(13.2m/s)^2 = 59590.08J

PE=(684kg)(9.8m/s^2)(5m) = 33516J

Total Energy= KE+PE = 93106.08

So. . .

Since we know that the velocity will be equal to zero when the car rolls to the highest point:

KE=0

PE=(684kg)(9.8m/s^2)(?m)

KE+PE=(684kg)(9.8m/s^2)(?m)+0= 93106.08 = 13.98 meters

My goodness that worked! Thanks for helping me in just right way!

 

[berkeman]

Glad to help. BTW, I like the way that you carry your units along in the calculations -- that is an important trick in calculations, and it will serve you well as you continue to learn more and more complex subjects. Good work.


EDIT -- I just noticed that there's a typo at the end of your calc -- you have 93106.08 = 13.98 meters, but you meant to show:

KE + PE = PE = 93106.08J --> H = 13.98 meters