Question

Two bicycle tires are set rolling with the same initial speed of 3.30 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 18.0 m ; the other is at 105 psi and goes a distance of 93.7 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 .

Answer 1

Answer:

At low pressure- $\mu_{k}=0.02315$

At high pressure- $\mu_{k}=0.00445$

Explanation:

Initial speed, $V_{i}=3.3 m/s$

Final speed, $V_{f}=3.3/2= 1.65 m/s$

Net horizontal force due to rolling friction $F_{net}=\mu_{k}$ mg where m is mass, g is acceleration due to gravity, $\mu_{k}$ is coefficient of rolling friction

From kinematic relation, $V_{f}^{2}- V_{i}^{2}=2ad$

For each tire,

$V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd$

Making $\mu_{k}$ the subject

$\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}$

Under low pressure of 40 Psi, d=18 m

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315$

Therefore, $\mu_{k}=0.02315$

At a pressure of 105 Psi, d=93.7

$\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445$

Therefore, $\mu_{k}=0.00445$