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Energy to move a car

  1. Jul 25, 2008 #1
    I'd like to calculate the energy required to move a 1000lb vehicle at 55mph for 1hr. I'm looking for a simple calculation, nothing with drag, friction or acceleration is required

    I expect this to be pretty simple, 1/2*m*v^2, to solve for the Kinetic Energy of the vehicle. However, how does 'time' come into play, to determine the energy needed over time?

  2. jcsd
  3. Jul 25, 2008 #2


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    If there is no drag or friction, once your car is going 55 mph it will need no energy to keep going at that speed.
  4. Jul 25, 2008 #3
    If there are no non-conservative forces acting on the car, then maintaining a constant velocity won't take any energy.
    Taking drag into account is relatively simple actually.
    Use the work-energy theorem [itex]W=\Delta E[/itex] and the definition of work
    [tex]W=\int_a^b \vec F \cdot \vec{dx}[/tex].
    The drag equation tells us that [itex]F_d=-\frac{1}{2}\rho A C_d v^2[/itex]. You can look this up on wikipedia for the details.
    Now you told us that the velocity needed to be a constant we can pull it out of the work integral and get (evaluating between x=0 and x=x),
    [tex]W=\frac{1}{2}\rho A C_d v^2 x[/tex]
    where x is the total distance traveled.
    Now [itex]v=\frac{dx}{dt}[/itex] and with v constant, [itex]x=vt[/itex] and so we see the total work done is proportional to [itex]v^3[/itex],
    [tex]W=\frac{1}{2}\rho A C_d v^3 t[/tex].
    For a reasonable car and under normal atmospheric conditions,
    [itex]C_dA\approx 7 m^2[/itex],
    [itex]\rho\approx 1.2 kg/m^3[/itex].
    So the work done after an amount of time t is numerically approximately (for normal cars, see the ACd product for cars on wikipedia),
    [tex]W\approx 5v^3t[/tex].
    For a car going 55 mph for 3600 s, the work is approximately [itex] 3 10^8 J[/itex]. That's quite a bit of energy!
    Hope that helps.
  5. Jul 31, 2008 #4
    I agree. lex_ee, your result will be overly-simplistic. You can find dynamometer data online for most cars. The data will give you rpms (or mph) and power in a more realistic setting--if you are given rpms, you will need appropriate car specs and data to convert to mph:

    The power can then be converted into the quantity that you desire.
  6. Aug 1, 2008 #5
    By Newton's first law once the car is moving at 55mph you will not need to add any more energy as you will not be accelerating it any more.


    [tex]\Delta{K} = \frac{1}{2}m(\Delta{v})^{2}[/tex] as [tex]\frac{dm}{dt}[/tex] is constant


    [tex]\approx\1.76 * 10^{6} [/tex] J
    Last edited: Aug 2, 2008
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