[SRMcDonald]

You are driving a 2500.0 kg car at a constant speed of 14.0 m/s along an icy, but straight and level road. While approaching a traffic light, it turns red. You slam on the brakes. Your wheels lock, the tires begin skidding, and the car slides to a halt in a distance of 25.0 m. What is the coefficient of sliding friction between your tires and the icy roadbed? Include a free body diagram.

Here is how I try to answer it:

For this question I only have the normal force in the equation F friction = F normal * coefficient, so I have to figure out the force of friction, which will be the net force located on the left of my free body diagram. I'm guessing the only force acting on the right would be inertia, which would be:

• F = m*a = (2500.0 kg)(-3.92 m/s/s) = -9800 N

I assume the force of friction on the locked tires and icy roadbed has to overcome the force of inertia plus the extra needed to accelerate at -3.92 m/s/s.

• F net = m*a = (2500.0 kg)(-3.92 m/s/s) = -9800 N

So the force of friction is twice the size of inertia pushing on the car as it tries to stop, making it -19 600 N. Now that I have this force, I can find the coefficient.

• Coefficient = F friction / F normal = (-19 600) / (24 500) = -0.8

Ignoring the negatives I attempted to make use of, I get 0.8 which seems a bit high, doesn't it? Where did I go wrong?

Here are my questions:
1) Did I do this wrong?
2) What would be the correct use of negative signs to display direction? I think I confused myself by trying.

[Capuchin]

Well, I have to go to bed, so I'll leave you this to think over: the immediate glaring problem is this "force due to inertia" you have.

Newton told us that a body remains at constant velocity unless a force acts on it. i.e. force creates an acceleration. I don't see where this forward force you have put on your free body diagram is coming from in the situation of the car braking and skidding, as it is not accelerating forwards.

This is a common misunderstanding. There is no forward force on a car at a constant velocity (ignoring friction). This "F Inertia" doesn't exist.

The only way a forward force could be created would be from the engine. But in a braking situation the engine is not doing anything. Think about it and you'll see that this F inertia is wrong.

I'll be back in the morning :)

[ebaines]

You did fine, right up to the factor of 2 multiplier. Your "F net" is correct - that's the force that the friction applies to the car to slow it down. So all you need is:

Coefficient = F net/F normal

Another way to do this problem is by determing the work the friction force needs to do to equal the initial kinetic energy of the car:



Here the force is applied by friction, which is:

[SRMcDonald]

Oh wow. Thank you guys for helping me. :)