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 93.5 m. Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g.A)What is the coefficient of rolling friction for the tire under low pressure?

B) What is the coefficient of rolling friction for the second one?

Answer 1

0.03219

0.00654

Step-by-step explanation:

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity =
`(u)/(2)=(4)/(2)=2\ m/s`

s = Displacement

g = Acceleration due to gravity = 9.81 m/s²

Acceleration

`a=-(f)/(m)\\\Rightarrow a=-(\mu mg)/(m)\\\Rightarrow a=-\mu g`

From equation of motion

`v^2-u^2=2as\\\Rightarrow v^2-u^2=2(-\mu g)s\\\Rightarrow \mu=(v^2-u^2)/(-2gs)\\\Rightarrow a=(2^2-4^2)/(2* -9.81* 19)\\\Rightarrow \mu=0.03219`

Coefficient of friction is 0.03219

`v^2-u^2=2as\\\Rightarrow v^2-u^2=2(-\mu g)s\\\Rightarrow \mu=(v^2-u^2)/(-2gs)\\\Rightarrow a=(2^2-4^2)/(2* -9.81* 93.5)\\\Rightarrow \mu=0.00654`

Coefficient of friction is 0.00654