Question
A bicycle and rider of total weight 815 N travel at a speed of 40.0 km/h.
The rider hits the breaks and the bicycle slides to a full stop in a distance of 8.5 m.
Ignore air resistance and use g=9.8 m/s2.
What is the coefficient of kinetic friction between the tires (rubber) and the road (pavement)?
The rider hits the breaks and the bicycle slides to a full stop in a distance of 8.5 m.
Ignore air resistance and use g=9.8 m/s2.
What is the coefficient of kinetic friction between the tires (rubber) and the road (pavement)?
Answers
N = total normal reaction = 815N
m = N/g = 83.16 kg
v0 = initial velocity = 40 km/h = 11.1 m/s
v1 = final velocity = 0
S = braking distance = 8.5 m
a=(v1²-v0²)/2S
μN = ma
Solve for μ
m = N/g = 83.16 kg
v0 = initial velocity = 40 km/h = 11.1 m/s
v1 = final velocity = 0
S = braking distance = 8.5 m
a=(v1²-v0²)/2S
μN = ma
Solve for μ
Thank you. I just wanted to confirm that this was the right way to do the question.
Just another clarification. In the FBD for this question, would the kinetic friction force be in the positive horizontal direction i.e. in the direction the rider was moving before he braked?
Just another clarification. In the FBD for this question, would the kinetic friction force be in the positive horizontal direction i.e. in the direction the rider was moving before he braked?
Since kinetic frictional force always opposes motion, by definition, it is opposite in direction to the movement of the bicycle. Also, this is how the bicycle is decelerated.