Calculating μs of a Car: Understanding Acceleration and Friction

[TAN6]

Find μs of the car.
- A Car takes 10 seconds to accelerate to 80km before the driver shifts the car into neutral.
- The cars mass = 1810kg
- The car coats 0.9km before coming to a complete stop.
- The total time = 86 seconds.

Here is the work I have done.

μs = ma/mg
to find a
V2 = V1 +at
22.2 = 0 +a10
a = 2.22m/s2

μs = 1810 * 2.22 / 1810 * 9.8
μs = 0.22


Does this make sense. This is the μs for the road right? How can I find it for the car? I figure I need to use the total distance it travel for something but I'm not sure how. Any help is welcome

Thanks!

 

[cepheid]

Here is the work I have done.

μs = ma/mg

I'm not sure I understand this. You're saying that the coefficient of friction multiplied by the normal force (which is just equal to the weight of the car in this case) is equal to some force, i.e.:

$$\mu_s mg = F$$

But why should this "F" be the net force on the car, ma?

to find a
V2 = V1 +at
22.2 = 0 +a10
a = 2.22m/s2

This looks reasonable, assuming constant acceleration.

Does this make sense. This is the μs for the road right? How can I find it for the car?

Umm...:uhh:...a frictional force is a force that opposes the motion of two surfaces in contact that are trying to slide relative to each other. So it only makes sense to speak of a coeffcient of friction between two surfaces. In this case, the only $\mu$ you could calculate would be the coefficient of friction between the tires and the road (i.e. the coefficient of friction for rubber on asphalt). It doesn't make sense to speak of mu only for the car, or only for the road. Are you sure this is the exact wording of the problem?EDIT: Ohhhh! I think I understand. In this case, the frictional force is the forward force of the road on the tires that causes the car to move forward instead of its tires just spinning in place, right? So that explains why you were setting the frictional force F equal to ma! It also explains why you were using the coefficient of static friction rather than kinetic friction (the two surfaces never actually slide relative to each other...unless the tires lose traction). Okay, in that case, I think you are on the right track, but you need to address the coasting phase of the trip too...not just the initial phase in which the engine is powering the car. During this phase, the situation is slightly different, right?

 

[m2003]

Your calculations for finding the acceleration and coefficient of static friction (μs) for the road are correct. However, to find the μs for the car, we need to consider the forces acting on the car during acceleration and deceleration.

During acceleration, the car experiences a forward force (F) from the engine and a backward force (Ff) from friction. The equation for acceleration (a) in terms of these forces is: a = (F - Ff)/m. We can rearrange this equation to solve for Ff: Ff = F - ma.

Now, during deceleration, the car experiences a backward force (Fb) from the brakes and a forward force (Ff) from friction. The equation for deceleration (a) in terms of these forces is: a = (Fb + Ff)/m. We can rearrange this equation to solve for Ff: Ff = a - Fb.

Combining these two equations, we can solve for the coefficient of static friction (μs) for the car: μs = Ff/F = (F - ma)/F = 1 - (ma/F).

Using the given information, we can calculate the forces acting on the car during acceleration and deceleration. The forward force (F) from the engine is equal to the car's mass (m) multiplied by its acceleration (a), which we calculated to be 2.22 m/s^2. Therefore, F = 1810 kg * 2.22 m/s^2 = 4012 N.

During deceleration, the backward force (Fb) from the brakes is equal to the car's mass (m) multiplied by its deceleration (a), which we can calculate using the given information about the car's speed and distance traveled. The average speed of the car during deceleration is 80 km/86 s = 0.93 m/s. The deceleration (a) can be calculated using the equation a = (Vf^2 - V0^2)/2d, where Vf is the final velocity (0 m/s), V0 is the initial velocity (0.93 m/s), and d is the distance traveled (0.9 km = 900 m). Plugging in these values, we get a = -0.0087 m/s^2.

Therefore, Fb = 1810 kg