Two bicycle tires are set rolling with the same initial speed of 3.90 m/s along
ID: 1424993 • Letter: T
Question
Two bicycle tires are set rolling with the same initial speed of 3.90 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 17.9 m ; the other is at 105 psi and goes a distance of 92.1 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 . Part A What is the coefficient of rolling friction r for the tire under low pressure? r = 3.25×102 SubmitHintsMy AnswersGive UpReview Part Correct Significant Figures Feedback: Your answer .0327 was either rounded differently or used a different number of significant figures than required for this part. Part B What is the coefficient of rolling friction r for the second tire (the one inflated to 105 psi)?
Explanation / Answer
r = (u2 - v2)/2gd where u2 - v2 = 3.92 - (3.9/2)2= 11.4075
for less pressure tyre => r = 11.4075/(2 x 9.8 x 17.9) = 0.0325148
for high pressure tyre => r = 11.4075/(2 x 9.8 x 92.1) = 0.0063194