Momentum Lab Help: Calculate Friction, Coefficient & Work

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The discussion revolves around a physics experiment involving two mini-carts that push away from each other, requiring calculations of friction, coefficient of friction, and work done by friction. Participants clarify that after the push, friction is the only force acting on the carts, which opposes their motion and can be calculated using the mass and acceleration. The correct formula for the coefficient of friction is emphasized as the force of friction divided by the normal force. Additionally, the total momentum of the system post-collision is defined as the sum of the momenta of both carts. The conversation concludes with the participant expressing gratitude for the assistance received.
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Homework Statement


The other day, I was asked to conduct an experiment involving momentum. Basically, I set two mini-carts bumper to bumper, and hit a switch that caused them to push away from each other and travel in opposite directions. We then had to repeat the experiment using different weights and such.

Anyway, after collecting the data, my physics teacher wants us to find the force of friction, coefficient of friction, and work done by friction. How do I approach these questions? I already know the carts’ Acceleration, Distance, Time, Initial Velocity, Final Velocity, Mass, Normal Force, and Momentum. Also, is acceleration supposed to be negative in these situations?

Homework Equations



Momentum = Mass * Velocity
Coefficient of Friction = Normal Force * Force of Frction
Friction = Force Applied from X Direction * (mass * acceleration)

The Attempt at a Solution



Fn - f = (mass x acceleration)
:18N - f = 1.8kg x -.106m/s^2 ?
::-f = -18
:::f = 18 N

Thanks for the help!
 
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No your calculation isn't correct. After the cart has been pushed, there is only one force acting on it. What do you think this is? This should answer your question about negative acceleration, and enable you to use Newton's second correctly.
 
cristo said:
No your calculation isn't correct. After the cart has been pushed, there is only one force acting on it. What do you think this is? This should answer your question about negative acceleration, and enable you to use Newton's second correctly.

Well, the only force acting on the carts after the push is Friction. I'm guessing since it's slowing the cart down, the force is negative - as it opposes forward progress. So, would friction be equal to (mass * -a)?
 
Correct. So, you now have the magnitude of friction. Can you calculate the coefficient and work done?
 
cristo said:
Correct. So, you now have the magnitude of friction. Can you calculate the coefficient and work done?

I believe so. All I have to do is plug the Force of Friction into the "Coefficient of Friction = Normal Force * Force of Friction, correct? :)
 
Well, the correct equation is force of friction=coefficient * normal force. But yes you just plus in the values you know.
 
Oh okay. Thanks for all the help, I really appreciate it!

Have a nice day. :-)
 
You're welcome!
 
Apparently, there is another question that asks, "What was the momentum of the system after collision?". What does my teacher mean when he says "momentum of the system"? Is it the average Momentum from both carts?
 
Last edited:
  • #10
Total momentum of the system will be the sum of the momenta of the two carts.
 
  • #11
Allrighty, thanks again!
 
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