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Homework Help: Work done against friction for a car with 125 KJ KE.

  1. Nov 5, 2013 #1
    "A 1200kg car as kinetic energy 125kJ at the bottom of a 20 degree slope. It rises to a height of 10m. Calculate work done against friction."

    Relevant equations:

    Work = force x distance in direction of that force

    Work = KE = 1/2 x m x v^2

    I'm not really sure where to start...tried several things but haven't got anywhere.

    For example:

    Work done = KE
    125000J = 1/2 x 1200 x v^2
    V^2 = 125000/600 = 208.3
    V = 14.4 m/s


    GPE at top of slope = mgh
    1200 x 9.81 x 10 = 117,720

    Energy lost = 125000 - 117720 = 7280 J

    But neither of these answer the question.

    Also tried

    Distance travelled = 10/sin20 = 29.24m

    So f = 125000/29.24 = 4274.9J buy apparently that's wrong.

    Someone please help? :-)
  2. jcsd
  3. Nov 5, 2013 #2

    Doc Al

    User Avatar

    Staff: Mentor

    This is the one you want.

    If that's not right: Are you sure it says "rises to a height " of 10m? Maybe they meant that it travels up the ramp a distance of 10 m?
  4. Nov 5, 2013 #3
    I tried it that way as well and none of the answers match up (it's multiple choice). Hm, perhaps I copied the answers/some of the question down wrong!
  5. Nov 5, 2013 #4
    So just to confirm, would the energy lost be the same as the work done against friction?
  6. Nov 5, 2013 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Yes, assuming friction is the only source of energy "loss".
  7. Nov 5, 2013 #6
    What are the choices?
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