# Kinetic Energy and Electric Car Power

1. Oct 26, 2012

### Jay_

If we consider the kinetic energy of say a car like Nissan Leaf moving at its top speed we have :

K.E. = 0.5 x M x V2

Given that M = 1521 kg, V (top speed) = 150 km/h ≈ 42 m/s

K.E. (@ top speed) = 1,341,522 ≈ 1.34 x 106 J ..... (1)

[1?] What connection does this value have with the battery energy or electric motor power?

[2?] When the car is in its top speed, this K.E. is the energy consumed per what unit of time? K.E. of a body is spoken of as belonging to the body that's moving, right? So in this case, the car's energy is (1) and since the stationary car gets this energy from the battery - does it keep getting this energy value constantly per second or per what unit of time?

The battery energy is WAY above this (which is okay, because as the source it has higher energy) = 24 kWh = 8.64 x 107 J ---> but greater as in 64.4 times? That may need comment. [3?]

The electric motor power is 80 kW = 8 x 104 W or J/s. And this seems less for the car to move at its top speed.

So what explains the mathematical or practical relation between these? [4?]

Last edited: Oct 26, 2012
2. Oct 26, 2012

### Staff: Mentor

Kinetic energy is not consumed, it is the quantity of energy required to accelerate the car to its cruising speed. That's why kinetic energy is energy and not power. By Newton's first law, no input power is required to keep an object moving unless there is something trying to slow it down.

So when a car is cruising, all of the energy being consumed (power) is going toward combating friction in the drivetrain and against the air.

3. Oct 26, 2012

### Packocrayons

The 24kW/h means that your car can put out 24kW for one hour. The amount of power required to accelerate the car to the top speed is 1341kW. Therefore accelerating at the rate you would be in order to attain that speed (unsure of what this rate would be? I'm going to say it's a second.), you could continue that acceleration for (24kW/h)/1341kW = 0.017 hours.
In other words you can accelerate up to that speed approx 60 times (0.017h= 1 min) assuming you have absolutely no resistance and at that point you would have used up all of your batter power.
Don't take any of what I said as fact, I really don't know how to solve the problem and I'm mostly guessing (with a grade 11 understanding of physics), so I hope something I wrote helps you answer.

4. Oct 27, 2012

### Jay_

Okay, and since the K.E. is the energy required to accelerate the car, it comes from the battery right? And in that sense the K.E. calculated is the energy required to accelerate the Nissan Leaf from 0 to top speed, correct? But this is the energy required to bring it to this speed in what amount of time?

Lets assume the electric motor is working its best to accelerate the car. So if the rated power is 80 kW, its peak can be assumed as 100 kW, which means its supplying 1 x 105 J/s.

And since car needs 1.34 x 106 J to reach top speed, does that mean it takes 1.34 x 106 / 105 = 13.4 seconds to reach top speed? Is this correct?

Now what is the connection between the battery energy and this K.E. of the car at a given velocity?

In this example for instance, lets say the electric motor works at its max (and theoretically doesn't wear off), so it gives 105 J/s. Does this mean that the car battery (energy = 8.64 x 107 J) will be completely discharged of energy after 8.64 x 107/105 = 864 seconds, or 14 minutes and 24 seconds, with the car going in top speed? Is that correct?

Another thing that may need comment is: In terms of physics principle we are talking about the wind acting against the car - so energy is continuously NEEDED to act against that force right? In what terms are we calculating this here above? Also, is FRICTION with the road the driving force? Because the wheels act in the opposite direction to the car's motion on the road.

Last edited: Oct 27, 2012
5. Oct 27, 2012

### Staff: Mentor

That's correct....or 1 kW for 24 hours, etc.
That is not correct. The calculation was for 1.34 kJ is energy, not power. You're making a similar mistake as the OP.

6. Oct 27, 2012

### Staff: Mentor

Correct.
Looks right, yes.
There is little or no connection. The battery is not sized based on top speed KE or any other KE, for that matter. It is sized to overcome the losses encountered while moving at a certain speed, for a certain amount of time. Wind resistance, friction in the drivetrain, etc.
No. As I said before, once moving at top speed, the kinetic energy is completely irrelevant. All of the power supplied by the battery at top speed goes toward overcoming wind resistance and frictional losses in the drivetrain.
Yes. And energy used continuously is called power. Watts and kW are power.
Are you asking how to calculate wind resistance? You can't without a complex computer model or wind tunnel.
The static friction of the road, yes. There is also rolling resistance, which is similar to a dynamic friction in the tires.

7. Oct 27, 2012

### Jay_

Hey Russ,

Now that would mean once accelerated, and in uniform top speed no energy would actually be required (in vacuum), but it would be required in case of the car because the wind and friction of the road are acting on it? Right?

And this energy being supplied per second is the electric motors power (rated at) 80 kW, meaning its rated to supply up to 8 x 104 J every second, so that the car doesn't get beat down by wind acting against it and the friction. Is that right?

-----

Coming back to the equation:

K.E. = 0.5 x M x V2 , -- this is the energy required to bring any body of mass 'M' to velocity 'V'. But wouldn't the energy required be different depending on how fast we bring the body to that velocity (i.e. depends on acceleration right)? But the equation above has nothing about acceleration or time. So that K.E. I calculated at first would be the energy required to bring it from rest to top speed in what time or acceleration?

Even the basic formula:

Energy = Force x distance, considers the acceleration, mass (in force), and the distance.

I attempted a solution as follows, with the derivation for K.E. (not sure if its right):

Energy = Force x distance => mass x acceleration x distance ......... (1)

Acceleration = (V-u)/t , and if starting from rest, u = 0, so:
Acceleration = V/t ......... (2)

D = ut + 0.5at2, and if starting from rest u = 0, so
D = 0.5at2
=> D = 0.5 x (V/t) x t2
=> Distance = 0.5 x V x t ............ (3)

Combining (1),(2) and (3) E = 0.5 x M x V2. However, the time associated with this, from (3) would be :

t = √[2D/V]. Would this be correct?

8. Oct 27, 2012

### willem2

If you ignore friction, the energy to bring the car to top speed doesn't depend on how fast you do it at all.

Your calculation only works for constant acceleration. An accelerating car will accelerate much faster at the start however. Because power = force * speed
= mass * acceleration * speed, if you accelerate with constant power, the
acceleration will have to go down if the speed goes up. The car will have to
change into a higher gear.
(at low speed, the force and thus the acceleration will be limited by the maximum friction of the wheels, and not all the power of the engine can be used)

I don't see how you get yourfinal result, or even where you made a mistake to get it.

The final result is wrong however, if you solve for t from (3), you get t = 2D/V, If this is true, your final result can't be.

9. Oct 27, 2012

### Staff: Mentor

Correct.
Correct.
No. If you accelerate slower, you use a lower power for a longer period of time, and it ends up the same. [qutoe] But the equation above has nothing about acceleration or time. [/quote] Exactly.
Any time, any acceleration.
Looks like you cancelled out something that you can't cancel-out: the acceleration is something arbitrarily chosen, so you can't just discard it and calculate a time.

10. Oct 27, 2012

### Jay_

My mistake.

It would give us:

t = √[2D/a] (not V as I typed earlier).

So that's fine, it means accelerating from rest to a given velocity at any rate would need a particular amount of energy. In this case, if a increased, time would decrease, but for any constant 'v' their effects would balance out for the energy I guess.

Last edited: Oct 27, 2012
11. Oct 27, 2012

### Jay_

Would it be accurate to say this: After the car has reached a constant velocity v, the battery energy is only used to fight the wind acting against the car, and that component of friction (between tires and the road) which doesn't act as the driving force?

I mean if the wheels of the car, are acting on the surface of the road are in a direction opposite to the motion of the car, it would mean that friction is the driving force, as said earlier. And yet, there would be some component of friction that is not is assisting in this motion I would guess. So, given that in vacuum we need no extra energy (after accelerating it) to keep the car in uniform motion, in this case to keep it in uniform motion the only energy needed would be to fight the wind resistance and a component of friction with the road. Is that correct?

12. Oct 28, 2012

### Staff: Mentor

Yes....though you should add-in the friction inside the car's drivetrain.
Yes.

13. Oct 28, 2012

### CWatters

Correct. Some energy is absorbed by the walls of the tyre as they flex where the tyre contacts the ground. This and other sources of friction total up to some value which is normally called the rolling resistance. You can sometimes reduce the rolling resistance by increasing the pressure in the tyres but there are disadvantages to that such as a worse ride and handling. Bicycle tyres used on smooth indoor tracks can run much higher pressures (80-120psi) than used on a road bike (typically half that).

14. Oct 28, 2012

### Jay_

Thanks guys :)