# Energy to move a car

1. Jul 25, 2008

### lex_ee

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

2. Jul 25, 2008

### mathman

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. Jul 25, 2008

### gamesguru

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 $W=\Delta E$ and the definition of work
$$W=\int_a^b \vec F \cdot \vec{dx}$$.
The drag equation tells us that $F_d=-\frac{1}{2}\rho A C_d v^2$. 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),
$$W=\frac{1}{2}\rho A C_d v^2 x$$
where x is the total distance traveled.
Now $v=\frac{dx}{dt}$ and with v constant, $x=vt$ and so we see the total work done is proportional to $v^3$,
$$W=\frac{1}{2}\rho A C_d v^3 t$$.
For a reasonable car and under normal atmospheric conditions,
$C_dA\approx 7 m^2$,
$\rho\approx 1.2 kg/m^3$.
So the work done after an amount of time t is numerically approximately (for normal cars, see the ACd product for cars on wikipedia),
$$W\approx 5v^3t$$.
For a car going 55 mph for 3600 s, the work is approximately $3 10^8 J$. That's quite a bit of energy!
Hope that helps.

4. Jul 31, 2008

### buffordboy23

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. Aug 1, 2008

### quark1005

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.

$$K=\frac{1}{2}mv^{2}$$

$$\Delta{K} = \frac{1}{2}m(\Delta{v})^{2}$$ as $$\frac{dm}{dt}$$ is constant

$$\approx\frac{1}{2}(\frac{1000}{2.2})(55*1.6)^{2}$$

$$\approx\1.76 * 10^{6}$$ J

Last edited: Aug 2, 2008