Two bicycle tires are set rolling with the same initial speed of 3.40 m/s along

Question

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 18.7 m; the other is at 105 psi and goes 92.0 m. Assume that the net horizontal force is due to rolling friction only. What is the coefficient of rolling friction for the tire under low pressure and the tire under higher pressure? (symbol is a funny looking 'u' with subscript 'r')

Explanation / Answer

Friction coefficient µ = F(rf)/F(normal) = ma/mg = a/g s = (v0^2-v1^2)/(2a) ==> a = (v0^2-v1^2)/(2s) At 40 psi, µ = (3.4^2-1.7^2)/(2*18.5*9.8) = 0.02391

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