Does coefficient of kinetic friction depend on speed?

[guyvsdcsniper]

Homework Statement

Does the coefficient of kinetic friction depend on speed? Explain, using your experimental data.

Relevant Equations

uKmg=m(-a)

In this part of the lab we pushed a block on a flat table and let it slide until it stopped. So it is decelerating with no force being applied to it while moving. In this case acceleration is negative. The only force acting on it is kinetic friction. Therefore I have come up with the following equation to solve for the coefficient of kinetic friction

.

Now the question in my lab ask does the coefficient depend on speed. I am inclined to answer no because according to my equation it depends on acceleration, but on the other hand acceleration is technically the change of velocity which is the change of speed. It related to speed in some way.

Am I overthinking this? Should the answer just be no it is not related to speed?

 

[haruspex]

according to my equation it depends on acceleration

Or does your equation says acceleration depends on the coefficient of friction?
Put it this way, how would you change the acceleration, keeping all else the same, in order to observe a different coefficient of friction?

 

[Steve4Physics]

Homework Statement:: Does the coefficient of kinetic friction depend on speed? Explain, using your experimental data.

Now the question in my lab ask does the coefficient depend on speed.

In the lab', did you measure $\mu_k$ for different initial speeds (keeping all other conditions the same)? The results of such measurements can be used to answer the question (for the range of speeds investigated).

 

[Chestermiller]

During your test, the velocity is changing. In order for the coefficient of kinetic friction to be independent of velocity, the acceleration would have to be constant. Is your distance vs time graph consistent with the acceleration being constant? If you are measuring velocity vs time, is the relationship linear?

 

[guyvsdcsniper]

Or does your equation says acceleration depends on the coefficient of friction?
Put it this way, how would you change the acceleration, keeping all else the same, in order to observe a different coefficient of friction?

Well, keeping all else to stay the same, a change in speed would change the the acceleration. So ultimately speed does play a factor in order to observe a different coefficient.

That makes sense to me. Does that sound right or am I still off?

 

[Chestermiller]

Well, keeping all else to stay the same, a change in speed would change the the acceleration. So ultimately speed does play a factor in order to observe a different coefficient.

That makes sense to me. Does that sound right or am I still off?

What does your data show?

 

[guyvsdcsniper]

What does your data show?

The trials with higher deceleration have a higher coefficient of kinetic friction. Deceleration being a change in velocity which is a change in speed. I guess there is my answer right there?

 

[Chestermiller]

Do you really think that the deceleration and coefficient of kinetic friction were constant during each individual trial?

 

[guyvsdcsniper]

Do you really think that the deceleration and coefficient of kinetic friction were constant during each individual trial?

Well when I think about it makes sense that they would be constant. The block gets pushed and then comes to a stop due to the coefficient of kinetic friction. That would be the only force acting on it in the horizontal direction.

 

[Steve4Physics]

@quittingthecult, could you tell us:
- what quantities you measured?
- how you calculated acceleration from these measurements?

 

[Chestermiller]

Well when I think about it makes sense that they would be constant. The block gets pushed and then comes to a stop due to the coefficient of kinetic friction. That would be the only force acting on it in the horizontal direction.

Really? Who says that the force acting on it in the horizontal direction is constant?

Suppose that $\mu_k=\mu_k(v)$ so that $$ma=m\frac{dv}{dt}=-mg\mu_k(v)$$ or$$\frac{dv}{dt}=-g\mu_k(v)$$so that $$\frac{dv}{\mu_k(v)}=-gdt$$

 

[guyvsdcsniper]

@quittingthecult, could you tell us:
- what quantities you measured?
- how you calculated acceleration from these measurements?

We used a motion detector to measure the velocity over a period of time. From that we were able to find the slope which was the acceleration. In this case the block was given an initial push and allowed to come to rest. The slope of the graph from push to rest was a linear slope downward which indicates the is decelerating.

So the only quantity I had was acceleration. With only that quantity I was left to find the coefficient of static friction and the Force of the kinetic friction. I used this formula

to find the coefficient. I then used that coefficient to find the kinetic friction force.Below are my results:

 

[Chestermiller]

Let’s see a graph of velocity vs time for an individual trial.

 

[guyvsdcsniper]

Let’s see a graph of velocity vs time for an individual trial.

Id have to get back to you on that. Not sure when I will get the data from my professor as this was all done on his computer.

 

[Steve4Physics]

We used a motion detector to measure the velocity over a period of time. From that we were able to find the slope which was the acceleration. In this case the block was given an initial push and allowed to come to rest. The slope of the graph from push to rest was a linear slope downward which indicates the is decelerating.

So the only quantity I had was acceleration. With only that quantity I was left to find the coefficient of static friction and the Force of the kinetic friction. I used this formula View attachment 279645 to find the coefficient. I then used that coefficient to find the kinetic friction force.Below are my results:
View attachment 279646

Your table gives average acceleration (and hence average $\mu_k$) during each run.

You have quite a wide spread of values (more than ±25%). Any suggestions why there’s such a large spread?

If $\mu_k$ depends on speed, then the frictional force, and hence acceleration, would change as the object slows down. This would result in acceleration changing during a run.

Do you have enough data to produce a velocity-time graph for each run? (If $\mu_k$ didn’t depend on velocity, what would the velocity-time graph look like?)

 

[Steve4Physics]

Let’s see a graph of velocity vs time for an individual trial.

You beat me to it.

 

[Chestermiller]

Do you have the data on the coefficient of friction as a function of the initial velocity of the block? How can you say there is no dependence on the velocity when you haven’t compared with even one piece of velocity data?