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

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SUMMARY

The discussion focuses on solving a physics problem involving a student and her backpack in an accelerating elevator. The backpack, with a mass of 8 kg, slides across a 3.2 m wide elevator floor while the elevator accelerates upward at 3.3 m/s². The coefficient of kinetic friction (Uk) is calculated using the formula Uk = (v²)/(2aL), resulting in a value of 0.21. Participants emphasize the importance of correctly calculating the normal force, which must account for both gravitational and acceleration forces acting on the backpack.

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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 a lot. 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?
 
Last edited:
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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