Posted by **shannon** on Thursday, July 3, 2008 at 10:19am.

A 250.0 kg roller coaster car has 20,000.0 J of potential energy at the top of the hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?

- physical science -
**drwls**, Thursday, July 3, 2008 at 2:44pm
The kinetic energy (1/2) M V^2 will equal the lost potential energy (20,000 J) at the bottom of the hill. This assumes that it had nearly zero kinetic energy at the top of the hill.

You know M, so solve for V.

V = sqrt(2*20,000/M)

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