# Power/Forces Help !

1. May 16, 2006

### atOnz

Power/Forces Help~~!

Hey there, first time using this forum. Have a question that I think I worked out correctly but am hoping to make sure I did (assignment).

A car must do 10kW to maintain a constant velocity of 25m/s on flat ground. How large are the frictional forces opposing it's motion?

Velocity = 25m/s
Power = 10 000W
Acceleration = 0 m/s^2

-First I converted kW into W: 10kW x1000W/1kW = 10 000W.

-Then I found the force applied (I think, please clarify):

P = Fv
10 000W = F(25m/s)
400N = F (applied?)

I asked my teacher a little on this question and he hinted at Force of Friction being 400N. But I don't understand why it could be that.

If the force applied and the force of friction are equal, doesn't that mean that the car is not moving? Please explain!

-Jay Matheson

2. May 16, 2006

### Hootenanny

Staff Emeritus
Hi Jay and welcome to PF,

Your working is absolutly correct. As for the balance forces, can you tell me what is Newton's first law?

~H

3. May 16, 2006

### atOnz

Law of Inertia - An object at rest or in uniform motion will remain at rest or in uniform motion unless acted on by an external force.

So the car will remain moving uniformly unless it is acted on by an external force. Friction being the external force.

F(applied) - F (friction) = 0N (net force)
400N - 400N = 0N (net force)

Then the car is moving uniformly, meaning that it is not at rest? So if there is no force acting on this car since friction and applied forces cancel each other out, then it is moving or not moving?

-Jay

4. May 16, 2006

### cscott

No, that only means there is no acceleration because there is no net force. The question states a "constant velocity of 25 m/s."

5. May 16, 2006

### Hootenanny

Staff Emeritus
Yep, that's correct. I would just like to add the work resultant into your law of inertia for clarification

"An object at rest or in uniform motion will remain at rest or in uniform motion unless acted on by a resultant external force."

Do you understand now?

~H

6. May 16, 2006

### atOnz

Just for curiosity, say that the force applied was greater than the force of friction, would that mean that the car was accelerating?

7. May 16, 2006

### Hootenanny

Staff Emeritus
Yes, by newton's second law;

$$\Sigma \vec{F} = m \vec{a}$$

There would be a resultant force, thus leading a and acceleration which is equal to the resultant force divided by the mass of the car.

~H

8. May 16, 2006

### atOnz

It's hard for me to get my head around that an object with no forces acting on it (because they cancel) can still be moving. Wouldn't there have to be a force to keep the car moving?

I think of it like this; If a book is sitting on a desk then F(normal) and force (gravity) are equal, meaning the book is not moving. Same as my situation except different vectors?

EDIT:

So I am still thinking (maybe overthinking? :P) but I have concluded this from what you wonderful people have said to help me:

F(applied) > F(friction) ~ Then there is acceleration.
F(applied) = F(friction) ~ Then there is uniform velocity (possibly at rest??)
F(applied) < F(friction) ~ Then there is no movement and no acceleration.

Cool?

Last edited: May 16, 2006
9. May 16, 2006

### Hootenanny

Staff Emeritus
Technically, it means the book it not accelerating. Consider this situation, a hockey player hits a puck across an ice rink. Once the puck is no longer in contact with the stick what are the force(s) acting?

~H

10. May 16, 2006

### atOnz

The only forces acting on the puck are F(normal), F(gravity), which cancel each other out. Also the F(friction) is gradually slowing down the puck. So the puck eventually stops. Meaning that the car in my question, which was constantly moving at a velocity of 25m/s will stop, So if the two forces are equal then the car will not stop until the force applied (engine) stops.

Haha I think I get it now! F(applied = F(friction) ~ uniform motion and will stay uniform unless some other force acts on it or the force applied < force of friction.

11. May 16, 2006

### Hootenanny

Staff Emeritus
Yep, you've got it. Just keep in mind net force and you'll have no worries

~H

12. May 16, 2006

### atOnz

Thank you very much for your help with this problem and for helping my better understand forces. It's much appreciated. I guess I was too focused on the whole: F(friction) = U(coeffcient of friction) x F(normal).

If I have more questions do I post them in this thread or start a new one?

13. May 16, 2006

### arildno

A new thread, for goodness sake.

14. May 16, 2006

### atOnz

Quick question:

In my actual question I am given 10.0kW and 25.0m/s. Would my finally answer of 400N be 4 x 10^2 since I am only given at most 2 sigfigs?

15. May 16, 2006

### Hootenanny

Staff Emeritus
Nope, it would be 400N and it appears to me that you are given three significant figures; 10.0 kW and 25.0 m.s-1. Also, in most cases it is acceptable to quote one more significant figure than was given.

~H

16. May 16, 2006

### atOnz

I thought 10.0 had only two significant figures since anything to the right of a non-zero digit is just a palce holder? I see why 25.0 is 3 but if by some chance 10.0 is 2 sig figs I should go with that many sigfigs, correct?

17. May 17, 2006

### Hootenanny

Staff Emeritus
Even if you consider 10.0 to have only two significant figures, you can still go to 3sf in your answer. It is perfectly acceptable to quote your answer to one additional significant figure than was given in the data; unless ofcourse you tutor / question has told you otherwise. For example, $\frac{5}{2} = 2.5 \neq 3$. Both the intial figures (5 & 2) were given to 1sf, however, the answer is given as 2.5 (2sf) not 3.

Eitherway, this is irrelavent, 10.0 has three significant figures.