Force equation problems

1. Aug 19, 2009

jeekeshen

Am getting really confuse with some force related equation..

Power = Workdone / Time

= Force X Speed

BUT THIS FORCE is it the one that makes the object move or the friction force( the one opposing the motion )

Because one of my homework makes me think that it's the frictional force, plzzz help

2. Aug 19, 2009

Born2bwire

Power = Force*Speed is correct. However, we can see that energy of a constant force applied over a distance d is F*d. If this force is applied over the distance d for a time t, then the average speed will be d/t. Thus, this you could say that P = F*v. More importantly, if we think of the force being applied over an infinitesimal distance dx, for an infinitesimal amount of time, dt, then we will find that P(t)=F(t)*(dx/dt)=F(t)*v(t). So this definition is true at any given time t.

3. Aug 19, 2009

tiny-tim

Welcome to PF!

Hi jeekeshen! Welcome to PF!
We usually talk about the power of a machine, so yes, that means using the force that makes the object move.

However, if the object isn't accelerating, that force will be the same as the friction force (well, ok, the opposite! ), and often the friction force is easier to calculate

I assume your homework question tells you only the speed the mass and the coefficient of kinetic friction ?

4. Aug 19, 2009

sganesh88

By definition, *any* force can do work. If the power is calculated for the friction force, it just gives the value of the quantity of energy dissipated by friction per second. We calculate the frictional power for determining the brake power of the engine-the one obtained at the flywheel.

5. Aug 19, 2009

jeekeshen

The real Question : The power of the car is 36.6 kw and speed is 32m/s
the mass of the car is 720 kg...

Calculate the magnitude of the external force opposing the motion of the car.....??

6. Aug 19, 2009

diazona

Well, in general you can have a power for each force - or more precisely, each force either applies or dissipates some power. So if you have a force that is pushing the car, like the force from the engine, you can calculate the power applied by that force, and if you have a force that is resisting the push, like friction, you can calculate the power dissipated by that force. An "applied" power will be positive, and a "dissipated" power will be negative; and if you add up the (positive) applied power from the engine and the (negative) dissipated power from the friction, you will get the net power exerted to move the car. That will be the same result you would get if you calculated the power exerted by the net force (engine minus friction) acting on the car.

In the question you posted, it looks like they give you the power exerted by the engine, and they want you to figure out the frictional force.

7. Aug 19, 2009

jeekeshen

Yes but i have its kinetic energy also... would this help to calculate the frictional force ???

8. Aug 20, 2009

tiny-tim

If the speed is constant, why would KE come into it?

9. Aug 20, 2009

Staff: Mentor

Since you're given the power and speed of the car, use that to figure out the effective force providing that power. Then ask: What's the net force on the car? (Is it accelerating?)

10. Aug 21, 2009

jeekeshen

No not accelerating.... the net force will perhaps be; using F= Ma where f is the resultant force, 720 X 9.8 = net force...

11. Aug 21, 2009

Staff: Mentor

If it's not accelerating, then the acceleration = 0. So what's the net force?

12. Aug 21, 2009

ideasrule

The car isn't in free fall!

13. Sep 1, 2009

jeekeshen

O yea acceleration is therefore zero WTF

14. Sep 1, 2009

Staff: Mentor

Right, the acceleration is zero and so is the net force. So the two forces acting on the car--the driving force and the opposing force--must be equal and opposite. The power equation will allow you to figure out the driving force.

15. Sep 1, 2009

Dweirdo

JUST for general knowledge,
is this question valid for 4 driving wheels( i mean, some of the cards(most) have two driving wheels(i hope that's who you call them) , in the front of the car where there the engine is )' but will it be valid?
or the question is for any force, not just friction with the ground, could be air drag as well?right?
thx