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?

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Answer:

0.03219

0.00654

Explanation:

t = Time taken

u = Initial velocity = 4 m/s

v = Final velocity = [tex]\dfrac{u}{2}=\dfrac{4}{2}=2\ m/s[/tex]

s = Displacement

g = Acceleration due to gravity = 9.81 m/s²

Acceleration

[tex]a=-\frac{f}{m}\\\Rightarrow a=-\frac{\mu mg}{m}\\\Rightarrow a=-\mu g[/tex]

From equation of motion

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2(-\mu g)s\\\Rightarrow \mu=\dfrac{v^2-u^2}{-2gs}\\\Rightarrow a=\dfrac{2^2-4^2}{2\times -9.81\times 19}\\\Rightarrow \mu=0.03219[/tex]

Coefficient of friction is 0.03219

[tex]v^2-u^2=2as\\\Rightarrow v^2-u^2=2(-\mu g)s\\\Rightarrow \mu=\dfrac{v^2-u^2}{-2gs}\\\Rightarrow a=\dfrac{2^2-4^2}{2\times -9.81\times 93.5}\\\Rightarrow \mu=0.00654[/tex]

Coefficient of friction is 0.00654