Posted by **Patricia** on Wednesday, June 6, 2012 at 1:44am.

An 800.0 kg roller coaster car is at rest at the top of a 95 m hill. It rolls down the first drop to a height of 31 m. When it travels to the top of the second hill, it is moving at 28 m/s. It then rolls down the second hill until it is at ground level.

What is the kinetic and potential energy at the top and bottom of each hill?

- PHYSICS -
**drwls**, Wednesday, June 6, 2012 at 2:55am
They presumably want to to assume total mechanical energy (kinetic + potential) is constant, even though that is not the case.

At the top of the 95 m hill, all of the energy is potential, and equals

M g H = 744,800 J. Use the height or velocity at the other location, and the total energy (748,800 J) , to determine kinetic and potential energies

- PHYSICS -
**Elena**, Wednesday, June 6, 2012 at 3:14am
1.

PE1 = mgh1 = 800•9.8•95 =744800 J.

KE1=0.

Total E1 = PE1+ KE1=744800 J.

2.

PE2 = 800•9.8•31=243040

PE1= PE2+KE2

KE2 = PE1- PE2 =

=744800 - 243040=501760 J.

Total E2 = PE2+ KE2=744800 J.

3.

KE3 =mv²/2= 800•(28)²/2 =313600 J.

PE3 =744800-313600 = 431200 J.

Total E3 = PE3+ KE3=744800 J.

4.

PE4=0.

KE4 = 744800 J.

Total E4 = PE4+ KE4=744800 J.

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