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

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In summary, the student kicks her backpack across the elevator floor, imparting to it speed v = 2.1 m/s. At time t = 2.2s, the backpack hits the opposite wall. The coefficient of kinetic friction between the backpack and the elevator floor is .21.
  • #1
cooltee13
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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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  • #2
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:
  • #3
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?
 
  • #4
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.
 
  • #5
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
 
  • #6
ya i figured it out now, couldn't have done it without your guys help though. Thanks a lot :biggrin:
 

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

1. What is the equation for calculating friction force?

The equation for calculating friction force is F = μN, where F is the friction force, μ is the coefficient of friction, and N is the normal force.

2. How do you determine the coefficient of friction?

The coefficient of friction can be determined by dividing the magnitude of the friction force by the magnitude of the normal force.

3. How does the mass of the student in the elevator affect the friction force?

The mass of the student does not directly affect the friction force. However, the normal force, which is affected by the mass, does impact the friction force as it is a component of the equation.

4. What is the significance of the width of the elevator in this problem?

The width of the elevator is used to calculate the normal force, which is a necessary component in the equation for determining friction force.

5. How can friction be reduced in this scenario?

Friction can be reduced in this scenario by decreasing the coefficient of friction, increasing the surface area in contact between the student and the elevator, or by applying a lubricant between the surfaces.

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