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.4,500 N

B.5,500 N

C. 3750 N

D.3125 N

Answer 1

In order to keep the car on the path, the force of friction must be the centripetal force. Use the definition of centripetal acceleration, a_c=v²/r, and substitute into Newton's Law F=ma to achieve the centripetal force.

F_f = F_c = ma_c

Now, make the substitution a_c=v²/r

F_f = m(v²/r)

m = mass (kg)

v = velocity (m/s)

r = radius (m)

F_f = (1250kg)(5m/s)²/10m

= (1250kg)(25m²/s²)/10m

= 3125 kg∙m/s²

= 3125 N