Car sliding down inclined plane.

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Homework Help Overview

The problem involves a car sliding down a 4.6 degree incline and coming to rest after traveling 30 meters, with a coefficient of kinetic friction of 0.45. Participants are exploring the dynamics of motion on an inclined plane, focusing on the forces acting on the car and the application of kinematic equations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants discuss using Newton's second law and the work-energy theorem to analyze the problem. Some suggest drawing free body diagrams to identify forces, while others express uncertainty about how to incorporate distance into their calculations.

Discussion Status

The discussion has progressed with participants sharing various approaches to set up the problem. Some have drawn diagrams and identified forces, while others are questioning the correct application of equations and the role of different forces. There is a mix of interpretations and attempts to clarify the relationships between forces and motion.

Contextual Notes

Participants are working under the constraints of not using formula sheets during tests, which influences their approach to memorizing and applying kinematic equations. There is also a focus on ensuring that forces are summed correctly according to their directions.

leroyjenkens
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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.
 
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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.)
 
You can also use the work energy theorem.
 
First draw a free body diagram and the forces acting on the car.
MrWarlock616 said:
You can also use the work energy theorem.
Using this eq you'll get v
Epot+Ekin-Fμ*s=0
 
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?
 
Write out the equation for the net downslope-directed force. For now, assume that the mass of the car is M.
 
gneill said:
Write out the equation for the net downslope-directed force. For now, assume that the mass of the car is M.

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.
 
leroyjenkens said:
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.

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:
:
:
 
gneill said:
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:
:
:
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
leroyjenkens said:
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?

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
gneill said:
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.
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
leroyjenkens said:
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.

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
gneill said:
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?

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
Yes. Bravo! :approve:
 
  • #15
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.
 

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