What Is the Coefficient of Friction for Fred's Rollerblades?

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NeomiXD
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Fred (60 kg) is rollerblading at a velocity of 25 km/h [E] when he sees a broken glass bottle on the path ahead. He brakes and is slowed to a velocity of 8 km/h [E] in 4.2 s. What is the coefficient of friction between Fred's rollerblades and the ground? (Hint: Find FN and Ff first.)

Given:

m = 60 kg
v1 = 25km/h = 6.94m/s [E]
v2 = 8km/h = 2.20m/s [E]
Δt = 4.2s
g = 9.8 N/kg

Required:

?

Solution:

Fg = mg
Fg = (60kg) (9.8 N/kg)
Fg = 588 N

FN = Fg
FN = 588 N
Ff ≤ µs FN
Ff ≤ µs (588 N)

??

I don't know what to do after.
µs is not given, so how do you solve this question?
 
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His initial kinetic energy must be equal to his final kinetic energy plus the amount of energy lost due to friction. So the amount of energy lost to friction is [itex]E_{ki} - E_{kf}[/itex] which is equal to [itex]mg\mu d[/itex]. Now you were given time instead of distance, but if you can assume constant deceleration, you could get d.
 
Hints: What's the skater's acceleration while applying the brakes? How much force is needed to create such an acceleration?
 
I would say you're looking for [itex]\mu k[/itex], as [itex]\mu s[/itex] is only applicable when masses have zero velocity. To solve for [itex]\mu s[/itex], you need to be told how much force is being applied to a body at rest when it just starts to budge.
 
So, if I'm solving for µk, how do I find Ff (the force of friction)? What formula do I have to use?
 
The force of friction is equal to the normal force multiplied by the kinetic coefficient of friction. The normal force is equal (in magnitude, but opposite in direction) to the force pulling the object down, namely, mg.
 
NeomiXD said:
Am I looking fo µs or µk?

If the wheels on the roller-blades do not skid on the ground, then you are working with µs. If the wheels are skidding, then you have µk .
 
Think kinematics. You have an initial velocity, a final velocity, and a time. Use this information to find the magnitude of the acceleration.