Question

Acceleration of gravity is 9.8 m/s

014 10.0 points To test the performance of its tires, a car travels along a perfectly flat (no banking) cir- cular track of radius 143 m. The car increases its speed at uniform rate of d lul at _ = 4.66 m/s2 dt until the tires start to skid. If the tires start to skid when the car reaches a speed of 26 m/s, what is the coefficient of atic friction between the tires and the road?

Answer 1

We know that friction force is given by coefficient of friction multiplied by normal force.

Here net force on the car is the frictional force.

F = mu N frictional force Normal force N = mg -(1) f = ma_net {a_t: tangential a_c: centripetal a_t = 4.66 m/s^2 a_c = v^2/r = (26)^2/143 = 4.73 m/s^2 a_net = squareroot a^2_c + a^2_t = 6.64 m/s^2 doublerightarrow mu = f/N = ma_net/mg mu = 6.64/9.8 = 0.677