Resistive force at constant velocity freewheeling down a slope

[ocfx]

Homework Statement

Ok so the problem is this: A car of a mass 960kg is free-wheeling down an incline (15 degrees to the horizontal) at a constant speed of 9.0 m s^-1
- Deduce that the average resistive force acting on the car is 2.4*10^3N

Homework Equations

F=ma I suppose, but it hasn't got me anywhere yet really any help would be appreciated.

The Attempt at a Solution

I calculated that the force should be F=mg = 960kg*9.81 = 9417.6N. Now I don't see how this will help in deducing the resistive force, but since the velocity is constant the resistive force should balance out any other force acting on it.

Any help will be appreciated. Thanks.

 

[willem2]

draw a free body diagram with all the forces in it

split all forces into components perpendicular and parallel to the slope. (the usual thing to for
problems with slopes)

 

[ocfx]

Could you or someone else clarify what forces would be involved despite friction and gravity. Also how will this help me deduce the resistant force acting on the car?

 

[Doc Al]

Could you or someone else clarify what forces would be involved despite friction and gravity.

The forces are gravity and some unknown resistive force (which you are trying to find).

Also how will this help me deduce the resistant force acting on the car?

What's the net force on the car?

 

[kuruman]

What is the acceleration of the car with the resistive forces?
What would be the acceleration of the car if there were no resistive forces?

 

[ocfx]

What is the acceleration of the car with the resistive forces?
What would be the acceleration of the car if there were no resistive forces?

The acceleration with resistive forces is 0 since the speed is constant but if there were no resistive forces the acceleration would be 9.81, right?!

 

[Doc Al]

The acceleration with resistive forces is 0 since the speed is constant but if there were no resistive forces the acceleration would be 9.81, right?!

That would be true if it were in free fall, but the car is constrained to go down an incline.

 

[ocfx]

That would be true if it were in free fall, but the car is constrained to go down an incline.

Right, so should I use trigonometry to prove it then that is sin 15 = 9417.6N/x ?
EDIT: that doesn't give me the right answer a little hint would be appreciated =D

 

[Doc Al]

Right, so should I use trigonometry to prove it then that is sin 15 = 9417.6N/x ?

That's incorrect.

What's the car's weight? What's the component of that weight parallel to the incline? (How do you find components parallel and perpendicular to an incline?)

 

[ocfx]

Well could you explain the concept, because I'm not exactly sure of it?

 

[Doc Al]

Read all about it: http://www.physicsclassroom.com/Class/vectors/u3l3e.cfm"