Car sliding down inclined plane.

1. Jan 12, 2013

leroyjenkens

A driver traveling down a 4.6 degree incline slams on his breaks, coming to rest 30m later. If the coeficient of kinetic friction between the tires and the road were 0.45, how fast was he initially traveling?

The only way I can think of solving this problem is to differentiate distance to get velocity. I'm not sure what to differentiate here. I don't have an equation. I guess I have to make my own? I have no idea how to do that.

2. Jan 12, 2013

Staff: Mentor

The acceleration is uniform; no need for differentiation, just a bit of kinematics.

Start with Newton's 2nd law. (Assume the wheels lock and the car slides.)

3. Jan 13, 2013

MrWarlock616

You can also use the work energy theorem.

4. Jan 13, 2013

lep11

First draw a free body diagram and the forces acting on the car.
Using this eq you'll get v
Epot+Ekin-Fμ*s=0

5. Jan 13, 2013

leroyjenkens

Ok so I drew the diagram and I have 3 forces acting on the car. I have the frictional force, the force of gravity, and the normal force.
To use F=ma, I need to add up all the forces and put them into the equation.
So in the x direction, the car has the force of kenetic friction, times the normal force, times cos4.6.
In the y direction, the car has the force of kenetic friction, times the normal force, times sin4.6.
I had those together and set them equal to ma.

I don't see how I can use this to include the distance of 30m, and then find the velocity. How do I do that with just F=ma?

6. Jan 13, 2013

Staff: Mentor

Write out the equation for the net downslope-directed force. For now, assume that the mass of the car is M.

7. Jan 13, 2013

leroyjenkens

I have U_k*N(cos4.6+sin4.6) = ma. I'm not sure if that's right, and I don't know how to begin involving the distance or how I'll get velocity out of this.

8. Jan 13, 2013

Staff: Mentor

No, that doesn't look right. What's N? Where's the mass of the car? Does gravity play a role?

Why not begin by listing the forces, there directions, and their formulas? I'll get you started:

Force due to gravity: M*g vertically downward
Downslope force due to gravity:
Normal force:
:
:

9. Jan 13, 2013

leroyjenkens

I drew the diagram and I have mg vertically downward.
Normal force is perp to the slope.
I'm not sure what the downslope force is. mgsin4.6?
Kinetic friction force up the slope.
Did I get them all?

10. Jan 13, 2013

Staff: Mentor

That'll do nicely. Assign formulas to each one. You formula for the downlsope force due to gravity is correct.

Then sum up the forces acting along (parallel to) the slope.

11. Jan 13, 2013

leroyjenkens

Using lecture notes to help me out on a different slope problem:

mgsin4.6 - u_k + N - mgcos4.6 = ma

N = mgcos4.6 so that one can go away I guess and leave me with just mgsin4.6 - u_k = ma? Since N is perp to the slope, it has nothing to do with the movement along the slope.

12. Jan 13, 2013

Staff: Mentor

Well, you should never do scalar addition of forces which are in different directions! Sum up the like components; the N and mgcos4.6 should not appear at all in the sum of components that are parallel to the slope.

Now, u_k is not a force; it's the coefficient of kinetic friction. How do you find the friction force?

13. Jan 13, 2013

leroyjenkens

Yeah, I meant to put i and j after the two terms, but forgot.

u_k is not a force, so I need to multiply it times the normal force?

So I get mgsin4.6 - u_kN = ma?

N = mgcos4.6, so I plug that in, divide both sides by m to eliminate it from both sides, plug in all the values I know to find a, which is -3.6 m/s^2.

Then use that information to plug into the kinematic equation v_f^2=v_i^2+2ad with the final velocity being 0, to find the initial velocity is 14.7 m/s?

14. Jan 13, 2013

Staff: Mentor

Yes. Bravo!

15. Jan 13, 2013

leroyjenkens

Thank you, and thanks for the help. We don't get to use formula sheets on tests, so I hope I can remember the kinematic equations.