Post a New Question

physic

posted by .

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?

  • physic -

    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

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

More Related Questions

Post a New Question