Question

Two bicycle tires are set rolling with the same initial speed of 4.0 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 18.2 m; the other is at 105 psi and goes a distance of 92.6 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 μr for the tire under low pressure?

Answer 1

Answer:

The coefficient of rolling friction will be "0.011".

Explanation:

The given values are:

Initial speed,

$v_i = 4.0 \ m/s$

then,

$v_f=\frac{4.0}{2}$

    $=2.0 \ m/s$

Distance,

s = 18.2 m

The acceleration of a bicycle will be:

⇒ $a=\frac{v_f^2-v_i^2}{2s}$

On substituting the given values, we get

⇒    $=\frac{(2.0)^2-(4.0)^2}{2\times 18.2}$

⇒    $=\frac{4-8}{37}$

⇒    $=\frac{-4}{37}$

⇒    $=0.108 \ m/s^2$

As we know,

⇒  $f=ma$

and,

⇒  $\mu_rmg=ma$

⇒       $\mu_r=\frac{a}{g}$

On substituting the values, we get

⇒       $=\frac{0.108}{9.8}$

⇒       $=0.011$