Two bicycle tires are set rolling with the same initial speed of 3.1 m/s 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 psi and goes a distance of 19.0 m m; the other is at 105 psi psi and goes a distance of 93.0 m m. Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g gg = 9.8 m / s 2 m/s2. Part A What is the coefficient of rolling friction μ r μrmu_r for the tire under low pressure?

Question

Brianna Willis

Physics

17 August, 05:09

Two bicycle tires are set rolling with the same initial speed of 3.1 m/s 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 psi and goes a distance of 19.0 m m; the other is at 105 psi psi and goes a distance of 93.0 m m. Assume that the net horizontal force is due to rolling friction only and take the free-fall acceleration to be g gg = 9.8 m / s 2 m/s2. Part A What is the coefficient of rolling friction μ r μrmu_r for the tire under low pressure?

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Camren Fox · Answer 1

From the formula

v² = u² + 2 a s, v is final velocity, u is initial velocity, a is acceleration and s is distance travelled.

u² / 4 = u² + 2 a x 19

- 3/4 u² = 2 a x 19

- (3 / 4) x 3.1² = 2 x 19 x a

a = -.18967 m / s²

deceleration due to friction = μg where g is acceleration due to gravity and μ is coefficient of friction.

a = μg

μ = a / g

=.18967 / 9.8

=.019.