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

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Johnny exerts a force of 790 N on Lucy when pushing her from rest for 1.0 second, while Lucy subsequently takes 8.0 seconds to come to a stop. The coefficient of kinetic friction is noted, but some participants argue that it does not significantly impact the scenario since roller blades are designed to roll. The discussion highlights the need to connect forces and acceleration despite the perceived lack of distance information. Participants suggest creating an unknown variable to simplify the calculations, indicating that the missing data may not affect the final answer. Overall, the focus is on understanding the dynamics of the push and subsequent motion.
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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.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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