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 17.9 m ; the other is at 105 psi and goes a distance of 92.0 m . Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g

Answer 1

Answer:

The coefficient of rolling friction for the tire under low pressure is 0.0342.

Explanation:

Two bicycle tires are set rolling with the same initial speed of 4.00 m/s

Final speed of both the bicycle, speed is reduced by half is measured, v = 2 m/s.

Here,

$u_kmg = ma\\\\a=\mu g$

Using third equation of motion as :

$v^2-u^2=2as\\\\v^2-u^2=2\mu gs\\\\\mu =\dfrac{v^2-u^2}{2gs}\\\\\mu =\dfrac{4^2-2^2}{2\times 9.8\times 17.9}\\\\\mu=0.0342$

So, the coefficient of rolling friction for the tire under low pressure is 0.0342.