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

At low pressure-
`\mu_(k)=0.02315`

At high pressure-
`\mu_(k)=0.00445`

Step-by-step 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`