Question

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along a long, straight road, and the distance each travels before its speed is reduced by half is measured. One tire is inflated to a pressure of 40 psi and goes a distance of 19.0 m ; the other is at 105 psi and goes a distance of 92.8 m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g = 9.80 m/s2 .

What is the coefficient of rolling friction μr for the tire under low pressure?

Answer 1

vi = 4 m/s
vf= 2 m/s
d = 19 m
use:
vf^2 = vi^2 + 2*a*d
2^2 = 4^2 + 2*a*19
a = (4-16) / (2*19)
= - 0.3158 m/s^2
This is the negative acceleration

Negative force causing this negative acceleration is rolling friction

a = -f/m
-0.3158 = - (μr *m*g) / m
-0.3158 = - μr*g
0.3158 = μr*9.8
μr = 0.032
Answer: 0.032