Electric car and battery problem

[ddobre]

Homework Statement

An electric car (mass 1560 kg) draws power from a set of 24 batteries. Each battery supplies 12 V, with storage 95 A·h. During a test of the vehicle, it is driven at exactly 45 km/h on a level road while experiencing a friction force of 240 N.
(a) What is the horsepower required during this test?
(b) Neglecting all other forms of energy loss (e.g. inefficiencies, acceleration), how many kilometers can the car travel before the batteries need to be recharged?

Homework Equations

P = A*V
1hp = 746w
P = F*v(velocity)

The Attempt at a Solution

The first thing I did was add up all the batteries voltages. Next I multiplied that value by the 95A*h rating to get P(watts) = 27360w*h. Next, I converted the speed from km/h to m/s: 12.5 m/s. Lastly, I tried finding the power of the frictional force on the car via P = F*v, (P = 240N * 12.5 m/s), and got 3000w. As a way of relating things, I wrote 27360w in terms of hp: 36.68hp. The part I am confused about is how I should be relating the frictional force to the horsepower required, and I'm also a little confused about the units of power I calculated for the frictional force. I have not thought about part b) yet. Any advice?

 

[haruspex]

I wrote 27360w in terms of hp

Why? You do not have a quantity 27360W.

how I should be relating the frictional force to the horsepower required

Is there any acceleration?

confused about the units of power I calculated for the frictional force

You wrote

P = 240N * 12.5 m/s

so it's Nm/s, also known as ?

 

[ddobre]

Why? You do not have a quantity 27360W.

Is there any acceleration?

You wrote

so it's Nm/s, also known as ?

Right, the quantity of have is 27360 w*h, not watts. There should be no acceleration on the car since it moves at a constant speed. I knew it was in terms of watts, but I am still unsure of how I should factor that in in terms of how much power is getting taking from the batteries simply due to friction.

 

[haruspex]

Right, the quantity of have is 27360 w*h, not watts. There should be no acceleration on the car since it moves at a constant speed. I knew it was in terms of watts, but I am still unsure of how I should factor that in in terms of how much power is getting taking from the batteries simply due to friction.

Think about the forces. What rate of work is friction doing against the car?

 

[ddobre]

Think about the forces. What rate of work is friction doing against the car?

That should be W = F*d, so that would be W = (240N) * (45000m) = 1.08x10^7 Joules

 

[haruspex]

That should be W = F*d, so that would be W = (240N) * (45000m) = 1.08x10^7 Joules

I wrote "rate of work", i.e. power.

 

[ddobre]

I wrote "rate of work", i.e. power.

Okay, so a force of 240N acting on the car as it travels at 12.5 m/s: P = 3000 Nm/s.

 

[haruspex]

Okay, so a force of 240N acting on the car as it travels at 12.5 m/s: P = 3000 Nm/s.

Right, so to maintain a steady speed, what power does the engine need to deliver?

 

[ddobre]

Right, so to maintain a steady speed, what power does the engine need to deliver?

If it provides 3000 Nm/s, then it would be equal to the frictional power. If the engine uses a larger amount of power than 3000 Nm/s, it would accelerate (?) So the engine needs to produce power equivalent to the rate of work of friction on it.

 

[haruspex]

If it provides 3000 Nm/s, then it would be equal to the frictional power. If the engine uses a larger amount of power than 3000 Nm/s, it would accelerate (?) So the engine needs to produce power equivalent to the rate of work of friction on it.

Right.

 

[ddobre]

Okay, that cleared some things up for me. For the second part I know I have to calculate the distance using the power of the batteries, but the units for the battery is w*h, so I'm thinking I could convert that to watts per second and then divide by the power needed by the engine to get the distance traveled. But I think that leaves the units in seconds, which is not quite what I'm looking for. But if I used V = d/t, then solve for d, then maybe this could work. I am getting a value of 410.4 km as the distance the car can travel.

 

[haruspex]

convert that to watts per second

You mean watt-seconds, = J.

410.4 km

Right.
Or more simply, work = F x d, where F=240N, work = 27360w*h.

 

[ddobre]

You mean watt-seconds, = J.

Right.
Or more simply, work = F x d, where F=240N, work = 27360w*h.

Ah yes that would be a much simpler solution than my method. This has been a lot of help, thanks. I don't get to talk to my professor or other students in the class very often.