Asked by Nikita

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)?

So do I use F= μ x mass of solid x g
and solve for μ

μ= F/(Mass of solid*g)
=F/815


Ok I feel as if its wrong, because I don't consider the speed or distance in this problem
Answered by MathMate
I agree, something is missing. The formula is correct if you knew the force, F.
In this case, the force is not know, but you can calculate the average force from energy considerations.
Work done by friction = F*D = 8.5F
using g=9.8 m/s², kinetic energy before hitting the <b>brakes</b>
= (1/2)mv²
= (1/2)(815/g)(40000/3600 m/s)²
Equate kinetic energy and work done and solve for F
F = 604 N

μ = F/(mg) = 604/815 = 0.74
Answered by risha
How did u arrive at F=604 N can you please describe/explain it more in detail.

thank you
Answered by Gharib
@risha,

equate (1/2)mv^2=8.5F
m and v are known, solve for F,

therefore F=604N
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