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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?

Respuesta :

Answer:

Explanation:

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 .