Finding the coefficient of kinetic friction

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SUMMARY

The discussion focuses on calculating the coefficient of kinetic friction (uk) for a cart on a linear track using photogate measurements. The instantaneous velocities measured were 1.05 m/s and 0.677 m/s, with a cart mass of 0.507 kg. The participants explored using the equations of motion, specifically vf = vo + at and Ff = ukFN, to derive uk, but faced challenges due to missing distance measurements between the photogates. Ultimately, they concluded that with the correct distance, acceleration could be calculated, allowing for the determination of the frictional force and uk.

PREREQUISITES
  • Understanding of kinematic equations, specifically d = vot + 1/2at²
  • Familiarity with Newton's second law, F = ma
  • Knowledge of frictional force equations, Ff = ukFN
  • Basic principles of motion and velocity measurement using photogates
NEXT STEPS
  • Learn how to derive acceleration using the equation vf² - vo² = 2ad
  • Research methods for accurately measuring distance in physics experiments
  • Explore the implications of varying distance on acceleration calculations
  • Study the relationship between kinetic energy and work done by friction
USEFUL FOR

Students in physics labs, educators teaching mechanics, and anyone interested in experimental physics and friction analysis.

rm_girl
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Homework Statement


For our lab we had to find the coefficient of kinetic friction for a cart on a linear track. We pushed the cart on the track and it went through two photogates which measured the time it spent in the photogate.
So through the 1st photogate = 0.0040s
2nd photogate = 0.0062s
There was a flag on the cart which was 0.42cm.
So we found the instantaneous velocity through the first gate as v=0.0042m/0.0040s = 1.05m/s
The instantaneous velocity through the second gate was v=.0042m/.0062s = 0.677m/s

The mass of the cart is = 0.507 kg

Homework Equations


vf = vo + at

F=ma

Ff = ukFN

FN = mg

The Attempt at a Solution


I was going to use vf = vo + at
find the acceleration as
a= vf - vo/t
but I don't know what the time is between the change instantaneous velocity.
I know if I get the acceleration I can plug into the F=ma, find the force and somehow get the frictional force from there.
 
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You will need one more piece of information. The distance between the gates will do. (But you didn't need the cart mass.)
 
Unfortunately we didn't measure the distance between the photogates either.
I only thought of getting the time/distance between the photogates AFTER the lab was done. :frown:
But I talked to my professor and apparently we can get the acceleration just from those two instantaneous velocity, unless he didn't understand our question when we asking him about the time issue.
 
rm_girl said:
apparently we can get the acceleration just from those two instantaneous velocity,
No chance. Suppose the distance between the gates was 1m. That would allow you to work out the acceleration, and would not contradict any of the information you provided. Now suppose it was 1km. This would still not contradict any known facts, but produce a rather different acceleration.
 
That's what I figured. Well, I can't go back and do the experiment again. So guess I'm screwed.
Thanks for the help though.
 
rm_girl said:
I can't go back and do the experiment again. So guess I'm screwed.
You can't even estimate it from memory?
For the purposes of the lab exercise, there is merit in at least demonstrating that you could have worked it out if you'd had all the measurements.
 
Do you have the time it took to go from one photogate to the other?
 
SammyS said:
Do you have the time it took to go from one photogate to the other?
No I didn't get that time.

I didn't set up the track and photogate, my lab partner did. And I very much highly doubt he remembers either.
I think my professor will accept if I explain it in my lab report what variables I'm missing, then it should be fine.
 
Just to make sure, before I start typing up my report.

If were to find the acceleration from the equation
a = vf-vi/Δt
Then I plug that into the F=ma
since frictional force is Ff = ukmg I would set
ma = ukmg
and find uk that way ?
 
  • #10
rm_girl said:
Just to make sure, before I start typing up my report.

If were to find the acceleration from the equation
a = vf-vi/Δt
Then I plug that into the F=ma
since frictional force is Ff = ukmg I would set
ma = ukmg
and find uk that way ?
Yes, but would you really have known Δt? How would you have measured it accurately? I think it's more reasonable that you would have noted the distance between the gates
 
  • #11
haruspex said:
I think it's more reasonable that you would have noted the distance between the gates

How would I find acceleration from the distance?
 
  • #12
rm_girl said:
How would I find acceleration from the distance?

There's a kinematic equation for that.
 
  • #13
which would be?
the only equation I know that involves distance and acceleration is
d = vot + 1/2at2
which has two unknowns in the equation.
 
  • #14
rm_girl said:
the only equation I know that involves distance and acceleration is
d = vot + 1/2at2
You haven't met vf2 - vo2 = 2ad?
If not, you could have derived it from the above together with any other kinematic equation by eliminating t.
 
  • #15
No, we never had to use that equation.
But I was just able to derive it using the equation I gave above. Thanks :smile:
 
  • #16
rm_girl said:
No, we never had to use that equation.
It's just force * distance = KE, with the masses canceled out.
 

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