Solve Car Acceleration: Force of Friction Homework

[anikam]

Homework Statement

A 1250 kg car traveling at 60.0 km/h comes to a sudden stop in 35 m. What is the coefficient of friction acting on the brakes?

HOW WOULD YOU FIND OUT THE ACCELERATION OF THE CAR?

Homework Equations

Fr= μN
Fn= ma

The Attempt at a Solution

I'm stumped on how to figure out the acceleration of the car as they only give you the mass, distance, and final velocity. Can anybdoy help me? Thanksss!

 

[Delphi51]

Do you have a list of formulas for constant accelerated motion?
There is one that does the job!

 

[anikam]

no i wasnt given a list, sorry.

 

[SammyS]

Homework Statement

A 1250 kg car traveling at 60.0 km/h comes to a sudden stop in 35 m. What is the coefficient of friction acting on the brakes?

HOW WOULD YOU FIND OUT THE ACCELERATION OF THE CAR?

Homework Equations

Fr= μN
Fn= ma

The Attempt at a Solution

I'm stumped on how to figure out the acceleration of the car as they only give you the mass, distance, and final velocity. Can anybody help me? Thanksss!

As Delphi51 said, you could use one of the kinematic equations.

However, you can work it out as follows.

Assuming uniform acceleration,you can easily determine the average velocity. (Do you know what the average velocity is during the time the car is braking?) Then from that you can find the time to go 35 meters. Then find the acceleration.

 

[asteriatic]

As a scientist, it is important to approach problems with a systematic and analytical mindset. In order to solve this problem, we need to use the equations for force of friction and Newton's second law of motion. First, let's identify the known values in the problem: mass (m = 1250 kg), initial velocity (vi = 60.0 km/h), final velocity (vf = 0 km/h), and distance (d = 35 m).

To solve for the acceleration, we can use the equation vf^2 = vi^2 + 2ad, where a is the acceleration. Substituting in the known values, we get 0^2 = (60.0 km/h)^2 + 2(a)(35 m). Solving for a, we get a = -0.483 m/s^2.

Now, we can use Newton's second law, Fn = ma, to solve for the normal force (Fn) acting on the car. The normal force is equal to the weight of the car (mg) in this case, so Fn = (1250 kg)(9.8 m/s^2) = 12250 N.

Finally, we can use the equation Fr = μFn to solve for the coefficient of friction (μ). Plugging in the known values, we get Fr = μ(12250 N) = (1250 kg)(-0.483 m/s^2). Solving for μ, we get μ = -0.000039, or approximately 0.004%.

In conclusion, the coefficient of friction acting on the brakes of the car is approximately 0.004%. It is important to note that this value may vary depending on factors such as the condition of the brakes and the surface of the road.