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

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Homework Help Overview

The problem involves determining the coefficient of friction between Fred's rollerblades and the ground as he decelerates from a higher velocity to a lower one over a specified time. The context includes concepts from kinematics and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy and friction, with some suggesting that energy lost due to friction can be expressed in terms of the coefficient of friction and distance. Others raise questions about whether to use static or kinetic coefficients of friction, and how to determine the necessary forces and acceleration.

Discussion Status

The discussion is active, with various approaches being explored. Participants are questioning the definitions of static versus kinetic friction and considering how to calculate acceleration and forces involved. Some hints have been provided regarding the use of kinematics to find acceleration, but no consensus has been reached on the method to find the coefficient of friction.

Contextual Notes

There is uncertainty regarding whether to use the static or kinetic coefficient of friction, as well as the implications of the rollerblades skidding or not. The problem is constrained by the information provided, specifically the use of time instead of distance.

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?
 
Am I looking fo µs or µk?
 
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.
 

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