A 1,250 kg Toyota Supra makes a right turn on a race track that has a radius of 10 meters. What is the force of friction on the Supra if the car makes the turn at a constant speed of 5 m/s?

Question

A. 0.66

B.0.25

C.0.33

D.0.15

Answer 1

Let F_f stand for friction

F_f = F_c = ma_c

Now plug in a_c=v²/r

F_f = m(v²/r)

F_f = 1250kg x (5m/s)² / 10 = 3125N (this is the force of friction)

The coefficient of friction can be found this way:

Fc = f = μsN = μsmg.

 μ = (5m/s)²/(10 meters x 9.8m/s²⁾ = 0.25