What's the Force Exerted? and how long it takes to stop

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Homework Help Overview

The problem involves two individuals, Johnny and Lucy, on roller blades, where Johnny applies a force to push Lucy, and the discussion centers around calculating the force exerted and the time it takes for both to come to rest. The context includes concepts of friction and motion, specifically relating to kinetic friction and acceleration.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between applied force and motion, questioning the relevance of the coefficient of friction in the context of rolling motion. Some participants suggest reconsidering the definitions and assumptions regarding forces and acceleration, while others express confusion about missing information and how to connect forces with acceleration.

Discussion Status

The discussion is ongoing, with participants providing insights and questioning the problem's wording and assumptions. Some guidance has been offered regarding the approach to the calculations, but there is no explicit consensus on the interpretation of the problem or the necessary information.

Contextual Notes

There is a noted concern about the wording of the question and the potential misunderstanding regarding the role of friction in the scenario. Participants mention the absence of distance as a factor in determining acceleration, which is part of the ongoing exploration of the problem.

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Homework Statement



Johnny, of mass 65 kg, and Lucy, of mass 45 kg, are facing each other on roller blades. The coefficient of kinetic friction between the roller blades and concrete surface is 0.20. When Johnny pushes Lucy from rest he applies a force for 1.0 s. Lucy then slows down to a stop in another 8.0 s. Calculate:
a. The applied force exerted by Johnny on Lucy. [790 N]
b. How long it takes Johnny to come to rest. [5.1s]

Homework Equations


F = ma
Vf = Vi + at
x(t) = Vi * t + 0.5*a*t^2

The Attempt at a Solution


Fk = (miu)*mass*9.81 = (0.2) (45Kg) (9.81 m/s^2) = 88.3 N
Vf = Vi + at = 0 + a(8s)
 
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Ouch. This is a very badly worded question. The coefficient of friction between the roller blades and the concrete has nothing to do with it. The roller blades will roll - that is what they are designed to do. The friction will do no work and therefore not slow Lucy at all.
What the question setter means (I hope) is rolling resistance. This is typically a result of friction at the axle of the wheel and/or deformation of the wheel as the load on it changes.
Putting that aside, you have calculated the force that will take Lucy from maximum speed (her speed when the push ends) to rest in 8 seconds. Now you want the force that goes the opposite way - from rest to maximum speed - in 1 second. Any thoughts?
You quote an equation involving an acceleration and two speeds. How can you connect that with forces?
 
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I tried connecting the two forces Fapp = - F john on lucy but I still need acceleration, which I can't figure out because there is no distance.
 
Hoyin said:
I tried connecting the two forces Fapp = - F john on lucy but I still need acceleration, which I can't figure out because there is no distance.
It often happens that it looks like there is not enough information, but it turns out that the information that seems to be missing doesn't affect the answer.
So just create an unknown for the missing data and run through the equations. You'll probably find the unknown cancels out.
In my previous reply I tried to give you a way of thinking about the question which would skip all that, but no matter.
 

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