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Homework Help: Energy - the ability to do work

  1. May 31, 2005 #1
    I'm fuzzy on what F actually is in the equation:

    W= F d

    is it only mg or the act of some force on an object

    I was absent when we started this unit at school and now I'm having trouble with the rest of the unit because of my uncertainty of this equation.

    Also, I'm having trouble with a specific question in my homework, I'll put down what I have, but it doesn't seem quite right.

    A 1550kg car is travelling down the highway at a constant speed of 33.3m/s
    a) If the car's engine is providing a power of 29, 500W how much force is it causing the ground to exert on the car?

    here's my answer/work:

    P=W/t = F v
    F= P/v = 29500/33.3 = ~885.8N

    now I'm unsure what to do from there, any help hints or suggestions would be helpful
  2. jcsd
  3. May 31, 2005 #2


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    The force in W=Fd is any force that is acting on anything that moves, provided the motion is in the direction of the force (or opposite). If the force and the motion are perpendicular, no work is done. If they are not perpendicular you need to find the component of the force in the direction of motion. Sometimes they are in the same direction and a simple product will do.

    Your calculation is fine. When force is constant, power is force times velocity, with the same considerations about direction described above.
  4. May 31, 2005 #3
    Thanks very much!

    If that first equation is fine then I guess my problem lies in the b part of the question.

    It asks how much energy must be removed from the car to slow it to 13.9m/s


    F = P/(Vf-vi) = 29500/(33.3-13.9) = 1003N

    but I'm fairly sure this isn't right because it would remove more energy than is present in a) part. (885.8)
  5. May 31, 2005 #4
    The answer is actually simpler, if you can find the kinetic energy of the car at 33m/s, and the kinetic energy at 13.9m/s then the difference between the two is the energy that needs to be removed.
  6. May 31, 2005 #5
    KE= 1/2 m (vf^2-vi^2) , right? At least for this equation
    Ke= .5(1550)(-33.3^2+13.9^2) = .5(1550)(-1108.89 + 193.21)
    KE= .5(1550)(-915.68) = -709652N

    That's quite the large number, that can't be the answer, can it?!!
    I'm quite unsure what I'm supposed to do with this number, or with the b) part of the equation for that matter.
    Last edited: May 31, 2005
  7. May 31, 2005 #6
    The first equation you have will give you the difference in kinetic energy. That is the answer to B. You've already solved A correctly with dan. It is alot of Joules, because a Joule is not a very large unit of energy. Its actually pretty small when dealing with very macro-scaled objects, such as a 1500kg car.
  8. May 31, 2005 #7


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    You are really talking about a change in kinetic energy, so it would be best write your left side as (KE_f - KE_i) rather than just KE, or if you use LaTex [itex]\Delta KE[/itex]. The units of energy are not Newtons. The units are Joules; a Joule is equal to one Newton*Meter or kg*m^2/s^2.
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