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A 250.0 kg roller coaster car has 20000 J of potential energy at the top of a hill. Neglecting frictional losses, what is the velocity of the car at the bottom of the hill?

The KE will equal the PE at the top

1/2 m v^2= 20000

solve for v.

Could you please put this in layman's terms?

I believe that this answer is 9.8 m/s am I right?

The kinetic energy (1/2 *mass*velocity squared) will equal the potential energy a the top

1/2 m v2 = 20000
The veloicity will be near 13 meters/second at the bottom

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