A roller coaster car (390 kg) moves from A (5.00 m above the ground) to B (19.0 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, ÄKE = KEf − KE0, from A to B?

Answer 1

The change in potential energy is
Δ PE = mgh2-mgh1 = 390•9.8•(19 -5) = 53508J
If the change in car's kinetic energy is ΔKE, then
ΔKE + 3.0•10^4 - 2.0•10^4 = Δ PE
ΔKE = 53508 -10000 = 43508 J