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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 .

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opudodennis

Answer:

At low pressure- [tex]\mu_{k}=0.02315[/tex]

At high pressure- [tex]\mu_{k}=0.00445[/tex]

Explanation:

Initial speed, [tex]V_{i}=3.3 m/s[/tex]

Final speed, [tex]V_{f}=3.3/2= 1.65 m/s[/tex]

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

From kinematic relation, [tex]V_{f}^{2}- V_{i}^{2}=2ad[/tex]

For each tire,

[tex]V_{f}^{2}- V_{i}^{2}=2\mu_{k}gd[/tex]

Making [tex]\mu_{k}[/tex] the subject

[tex]\mu_{k}=\frac {V_{f}^{2}- V_{i}^{2}}{2gd}[/tex]

Under low pressure of 40 Psi, d=18 m

[tex]\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*18}=-0.02315[/tex]

Therefore, [tex]\mu_{k}=0.02315[/tex]

At a pressure of 105 Psi, d=93.7

[tex]\mu_{k}=\frac {1.65^{2}- 3.3^{2}}{2*9.8*93.7}=-0.00445[/tex]

Therefore, [tex]\mu_{k}=0.00445[/tex]

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