Posted by **sarah** on Saturday, May 31, 2008 at 7:06pm.

A rollercoaster car of mass 600.0 kg is moving 2.4m/s at the top of a peak that is 25m off the ground. The car moves down slope and up the next hill. Assume that it loses 20,000J of energy by the next time it gets to the top of the next peak due to friction forces. If the speed of the car must be at a minimum of 2.0m/s to keep it moving on to the next slope, what is the maximum height allowed for the 2nd peak.

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**drwls**, Saturday, May 31, 2008 at 7:26pm
The initial total(kinetic + potential) energy, with respect to ground level zero P.E., is

(1/2) M V^2 + M g H = 1728 J + 147,150 J = 148,778 J. If it loses 20,000 J before reaching the next peak, and has 2.0 m/s velocity there, then

128,778 = (1/2) M V^2 + M g H'

= 1200 + M g H'

Solve for the new maximum height, H'. I get H' = 21.7 m.

I used 9.81 m/s^2 for g. No guarantees here.. check my assumptions and calculations.

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