Solve Friction Problem: Student in Elevator, Mass 8kg, Width 3.2m

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A student seeks assistance with a physics problem involving a backpack sliding in an accelerating elevator. The elevator accelerates upward at 3.3 m/s², and the backpack has a mass of 8 kg. After calculating the normal force and using kinematic equations, the student initially miscalculated the coefficient of kinetic friction. With guidance, they corrected their approach by accounting for both the weight and the elevator's acceleration. Ultimately, the correct coefficient of kinetic friction was determined to be approximately 0.0447.
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[SOLVED] Friction Problem. need help!

Hey guys, I am new on here. I have a test coming up and this was an example from an old test. I was wondering if someone could help me out and let me know the steps involved/ how to solve it. It would help me out ALOT. thanks

A student stands in an elevator that is continuously accelerating upward with acceleration a = 3.3 m/s^2. Her backpack (mass m = 8.0kg) is sitting on the floor next to the wall. The width of the elevator car is L = 3.2 m. The student gives her backpack a quick kick at t = 0, imparting to it speed v = 2.1 m/s, and making it slide across the elevator floor. At time t = 2.2s, the backpack hits the opposite wall. Find the coefficient of kinetic friction between the backpack and the elevator floor.
 
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Welcome to PF,

For future reference, we have Science Education forums (see my signature) for bookwork questions. Now, for your actual question: Can you start by calculating the normal force exerted on the back-pack?
 
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ahh ok, thanks for the tip. Ill post there next time.

Heres what I've found so far:

Uk(coefficient of kinetic friction)N=Fk ,

using N=mg, i got n=78.4 however because the elevator is moving up i found n= 26.4

from there i used the equation Vxf^2 = Vxi^2 + 2ax(Xf-Xi)
I plugged in -Uka for ax ( based on a given equation) therefor,
0 = Vxi^2-2(Uka)Xf
(Vxi)^2/2AXf = Uk

Uk = (2.1 m/s)^2/2(3.3m/s^2)(3.2) => .21 = Uk

Im pretty sure i made a few mistakes in there. Does that look right?
 
Your problem stems from here,
cooltee13 said:
using N=mg, i got n=78.4 however because the elevator is moving up i found n= 26.4
I suggest that you draw yourself a FBD and mark on all the forces. You should note that the normal force must account for both the weight and the acceleration of the back-pack.
 
Well u must compute the negative acceleration of the backpack. This should be a=f*(9.81+3.3). Then insert this to kinetic equation v*t - 0.5*a*t^2 = s, u know everything except f, so 2.1*2.2 - 0.5*f*(9.81+3.3)*2.2^2 = 3.2 The result = 0.0447
 
ya i figured it out now, couldn't have done it without your guys help though. Thanks a lot :biggrin:
 
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