Two bicycle tires are set rolling with the same initial speed of 3.40 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 18.5 m ; the other is at 105 psi and goes a distance of 92.3 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 for the tire under low pressure? for the tire of higher pressure?

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aristocles aristocles · Answer 1 · 2019-11-11T01:48:55+01:00

Answer:

for low pressure tyre we have

[tex]\mu_1 = 0.024 [/tex]

Now similarly for high pressure tyre

[tex]\mu_2 = 0.0048 [/tex]

Explanation:

As we know that both the bicycle tyres are set into rolling motion with initial speed given as

[tex]v_i = 3.40 m/s[/tex]

now their speed becomes half of the initial speed after travelling some given distance

so we can use kinematics to find the friction coefficient

first for low pressure tyre we have

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex](1.70)^2 - (3.40)^2 = 2(-\mu_1g)(18.5)[/tex]

[tex]\mu_1 = 0.024 [/tex]

Now similarly for high pressure tyre

[tex]v_f^2 - v_i^2 = 2a d[/tex]

[tex](1.70)^2 - (3.40)^2 = 2(-\mu_2g)(92.3)[/tex]

[tex]\mu_2 = 0.0048 [/tex]