Energy to move a car

  • Thread starter lex_ee
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  • #1
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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?

Thanks
 

Answers and Replies

  • #2
mathman
Science Advisor
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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.
 
  • #3
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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?

Thanks
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.
 
  • #4
548
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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.

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:
http://www.type2.com/library/misc/calcspd.htm

The power can then be converted into the quantity that you desire.
 
  • #5
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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?

Thanks

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]K=\frac{1}{2}mv^{2}[/tex]

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

[tex]\approx\frac{1}{2}(\frac{1000}{2.2})(55*1.6)^{2}[/tex]

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

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