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Homework Help: Car Work and Power Calculation

  1. Feb 4, 2009 #1
    1. The problem statement, all variables and given/known data

    A car of mass 840.0 kg accelerates away from an intersection on a horizontal road. When the car speed is 37.4 km/hr (10.4 m/s), the net power which the engine supplies is 4300.0 W (in addition to the extra power required to make up for air resistance and friction). Calculate the acceleration of the car at that time.

    2. Relevant equations




    3. The attempt at a solution

    I have tried numerous ways of solving this problem and remain unsuccessful. My closest attempt was using the equation Power=Kinetic Energy/Time, I used the velocity as Kinetic Energy to find the time and plugged that into the Kinematic equation Vf=Vi+a*d but that did not work either. Please help me!
  2. jcsd
  3. Feb 4, 2009 #2


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    Do you know some calculus?
    It looks like a calculus problem to me.
    You can't use Power = KE/time: consider the situation where the car is moving at zero acceleration and a large speed. The power required ("extra" not counted as stated in problem) is zero but the KE is large.

    Instead, you should use Power = dE/dt, the derivative of kinetic energy with respect to time. The differentiation is not difficult, but does involve the chain rule.
  4. Feb 5, 2009 #3
    yes I do know calculus, I am in Calc 2 but I still don't follow.
  5. Feb 5, 2009 #4
    M= 840.0 kg
    P= 4300.0 W
    v= 10.4 m/s

    s = v*dt [you are working with power (which is watt per second). what should dt be?]

    gives s

    p = w/dt

    gives w

    w = f*s

    gives f

    f = m*a
  6. Feb 5, 2009 #5
    how am I supposed to find dt when im not given a time
  7. Feb 5, 2009 #6


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    As I recall power is equal to force times velocity.

    Units of Watts = N-m/s

    P = F*V

    So the instantaneous power is given as 4300 w and the velocity is 10.4 m/s.

    That makes F = 4300/10.4 doesn't it?

    And you have the mass, so ...
  8. Feb 5, 2009 #7
    i thought that power was equal to F*V but if you do the math it delivers units of Kg*m/s and not Kg*m/s2. maybe im using the wrong velocity?
  9. Feb 5, 2009 #8
    Watt = J/s = kgm2/s3 = Nm/s

    So it works out just fine. :)
  10. Feb 5, 2009 #9


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    Oh, clever to remember P = Fv !

    Just for interest, I was thinking of P = dE/dt = d/dt(.5mv^2) = .5m*2v*dv/dt = mva
    which is the same as P = Fv.
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