# Help on force, work, velocity and friction?

1. Oct 8, 2011

### Tuvshee

1. The problem statement, all variables and given/known data
It says: A car moves at a constant speed of 5.0ms-1 along a 6.0km road against frictional force of 150N per meter. How much work has to be done by the car to maintain its constant speed?

2. Relevant equations
W = F*d

3. The attempt at a solution
So I just did 150N * 6000m = 900000J = 900kJ

But what about the 5ms-1? Is that important? I'm guessing it is since it is mentioned, but I don't know how to apply it. And the answer sheet says 9000kJ (and not 900kJ). Please help me? I'm confused, thank you.

2. Oct 8, 2011

### rock.freak667

The frictional force is 150 N/m so if you multiply that by the 6000 m you will get the frictional force experienced throughout the entire 6000 m. In order to maintain the 5 m/s you'd need to get rate of work that needs to be done to overcome the frictional force.

3. Oct 8, 2011

### Tuvshee

Ok I get that. But if the unit of frictional force is N/m, then it's not a force anymore, is it? Because unit for force is N and not N/m. Anyways: force would be:
150N/m * 6000m = 900000N.

v = 5m/s
N = 900000N

Then? Which formula should I use?

4. Oct 8, 2011

### Villyer

Have you tried to approach the problem using power?
Whenever I see velocity mentioned, I think of power.

5. Oct 8, 2011

### rock.freak667

They didn't give you the force for the entire length, they just gave you the force per unit length, so it is still a force just defined per unit length.

Normally, I'd think they were asking for power rather than the work done.

6. Oct 9, 2011

### Tuvshee

Oh ok :) well they were asking for the work done...oh and correction. The answer sheet says 900J. But it's still different than my result (900kJ). And I'm still stuck with no clue. :( And I don't think power is involved, because we haven't covered the topic yet.