Two bicycle tires are set rolling with the same initial speed of 4.00 m/s along
ID: 1425807 • Letter: T
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 r for the tire under low pressure?
Explanation / Answer
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 = - (r *m*g) / m
-0.3158 = - r*g
0.3158 = r*9.8
r = 0.032
Answer: 0.032