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

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SUMMARY

The discussion focuses on calculating the force exerted by Johnny on Lucy during a push while they are on roller blades, with specific parameters provided. Johnny applies a force of 790 N to Lucy, who has a mass of 45 kg, and comes to a stop in 8 seconds due to kinetic friction with a coefficient of 0.20. The time it takes for Johnny to come to rest after pushing Lucy is determined to be 5.1 seconds. The conversation highlights the importance of understanding the role of friction and acceleration in motion equations.

PREREQUISITES
  • Understanding of Newton's second law (F = ma)
  • Knowledge of kinematic equations (Vf = Vi + at)
  • Familiarity with the concept of kinetic friction and its coefficient
  • Basic grasp of roller blade dynamics and rolling resistance
NEXT STEPS
  • Study the effects of rolling resistance on motion in physics
  • Learn how to apply Newton's laws to real-world scenarios
  • Explore the relationship between force, mass, and acceleration in different contexts
  • Investigate the impact of friction coefficients on various surfaces
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Students studying physics, particularly those focusing on mechanics, as well as educators looking for practical examples of force and motion concepts.

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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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