A roller coaster (375. kg) moves from A (5.00 m above the ground) to B (28.0 m above the ground). Two nonconservative forces are present: friction does -2.00 104 J of work on the car, and a chain mechanism does +3.00 104 J of work to help the car up a long climb. What is the change in the car's kinetic energy, ΔKE = KEf - KE0, from A to B?

asked by Anonymous

13 years ago

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1 answer

Potential Energy increase going from A to B = 375*23*9.8 = 8.453 * 10^4 J

PE gain + KE gain = Chain work -Friction loss

answered by drwls

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Question

A roller coaster (375. kg) moves from A (5.00 m above the ground) to B (28.0 m above the ground). Two nonconservative forces are present: friction does -2.00 104 J of work on the car, and a chain mechanism does +3.00 104 J of work to help the car up a long climb. What is the change in the car's kinetic energy, ΔKE = KEf - KE0, from A to B?

1 answer

Potential Energy increase going from A to B = 375*23*9.8 = 8.453 * 10^4 J

PE gain + KE gain = Chain work -Friction loss

drwls · Answer

Potential Energy increase going from A to B = 375*23*9.8 = 8.453 * 10^4 J PE gain + KE gain = Chain work -Friction loss