Coefficient of friction/conservation of energy problem

[poodle123]

Homework Statement

An object is launched along the incline of angle 30 degrees with horizontal from its bottom level with initial velocity 6.5 m/s. It reaches height 2.1 m, comes to momentarily stop and slides back. When it comes back to initial point it has velocity 2.5 m/s. Find coefficient of friction between object and an incline.

Relevant Equations

coefficient of friction, conservation of energy

Not sure if I did something wrong but I am getting two different coefficients of friction for each velocity.

 

[poodle123]

Not sure if I did something wrong but I am getting two different coefficients of friction for each velocity.

 

[kuruman]

If you don't show us what you did, we cannot tell you what you did wrong.

 

[poodle123]

Don't worry I was able to figure it out just now. Thank you though.

 

[kuruman]

Don't worry I was able to figure it out just now. Thank you though.

I'm glad you figured it out. For future reference please remember to show your work as best as you can when you seek help.

 

[haruspex]

Homework Statement:: An object is launched along the incline of angle 30 degrees with horizontal from its bottom level with initial velocity 6.5 m/s. It reaches height 2.1 m, comes to momentarily stop and slides back. When it comes back to initial point it has velocity 2.5 m/s. Find coefficient of friction between object and an incline.
Relevant Equations:: coefficient of friction, conservation of energy

Not sure if I did something wrong but I am getting two different coefficients of friction for each velocity.

Yes, there is too much information, so it's likely you can get more than one answer...
... unless we are wrong to assume this is at the surface of the Earth.

What approaches did you use and what answers do you get?

 

[poodle123]

Hi thank you so much for replying! I was able to come to the answer this was how it was done!

 

[poodle123]

Here was how it was supposed to be done.
Hi thank you so much for replying! I was able to come to the answer this was how it was done!
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[haruspex]

Hi thank you so much for replying! I was able to come to the answer this was how it was done!
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You have got an answer by ignoring some information, so that does not resolve your issue.
You can use that it reaches a height of 2.1m (ignoring what happens later), or that starting from rest at that height it reaches the given final speed, or, as you have done here, considered the loss of KE over the whole trip. These will all give different answers unless the problem setter has been very careful to make sure the data are consistent.

Edit: using the first two of the above approaches, I get 0.015 and 0.49. If I resolve the discrepancy by letting g be unknown I get that g is 5.77m/s² and mu is 0.43.

 

[poodle123]

You have got an answer by ignoring some information, so that does not resolve your issue.
You can use that it reaches a height of 2.1m (ignoring what happens later), or that starting from rest at that height it reaches the given final speed, or, as you have done here, considered the loss of KE over the whole trip. These will all give different answers unless the problem setter has been very careful to make sure the data are consistent.

Edit: using the first two of the above approaches, I get 0.015 and 0.49. If I resolve the discrepancy by letting g be unknown I get that g is 5.77m/s² and mu is 0.43.

I got 0.015 and 0.49 for mu at first, but after inputting the answers into the homework software, it said it was wrong. But I tried 0.2525 for mu, and it worked. I am not really sure why there is a discrepancy between the coefficient of friction over the whole trip, the coefficient of friction going up, and the coefficient of friction going down. I mean technically, shouldn't the coefficient if friction not change since the surface is constant? I mean I don't think coefficient of friction is dependent on gravitational acceleration, unless I am wrong (I was just under the assumption that gravitational acceleration didn't affect the coefficient of friction). Another thing you can assume in this problem though is that gravitational acceleration is 9.8 m/s^2. I known it isn't mentioned explicitly but it is assumed that this "experiment" is taking place on earth.

 

[haruspex]

I got 0.015 and 0.49 for mu at first, but after inputting the answers into the homework software, it said it was wrong. But I tried 0.2525 for mu, and it worked. I am not really sure why there is a discrepancy between the coefficient of friction over the whole trip, the coefficient of friction going up, and the coefficient of friction going down. I mean technically, shouldn't the coefficient if friction not change since the surface is constant? I mean I don't think coefficient of friction is dependent on gravitational acceleration, unless I am wrong (I was just under the assumption that gravitational acceleration didn't affect the coefficient of friction). Another thing you can assume in this problem though is that gravitational acceleration is 9.8 m/s^2. I known it isn't mentioned explicitly but it is assumed that this "experiment" is taking place on earth.

As I posted, the issue is that the problem statement gives inconsistent facts. You can get all sorts of different answers depending on which facts you trust.
For the uphill journey, ##\mu=(\frac{v_i^2}{2gh}-1)\tan(\theta)##.
For the downhill journey, ##\mu=(1-\frac{v_f^2}{2gh})\tan(\theta)##.
For that to be consistent, ##4gh=v_i^2+v_f^2##.
The four values you are given for those variables disagree.

 

[poodle123]

I see, thank you so much for your help! I appreciate it.

 

[gmax137]

Thanks to @poodle123 and @haruspex for working this through. I sometimes try to work these problems myself, just for fun. On this one, I too, got three answers (mu uphill, mu downhill, and mu up&down). Made me wonder if I was doing something wrong there. Glad to find out I wasn't way off base.

I never thought of changing the g value to "cure" the discrepancy.

 

[kuruman]

Perhaps a better way to resolve the discrepancy is to leave ##g## alone and make the angle of the incline also unknown in which case the distance ##s## traveled along the incline must be given instead of the vertical displacement ##h##. One has to solve a system of two equations and two unknowns: (a) the angle of the incline and (b) the coefficient of kinetic friction. The relevant equations are $$2(g\sin\theta+\mu_kg\cos\theta)s=v_i^2;~~~2(g\sin\theta-\mu_kg\cos\theta)s=v_f^2$$ from which we obtain the unknowns by adding and subtracting,$$\sin\theta=\frac{v_i^2+v_f^2}{4gs};~~~\mu_k=\frac{v_i^2-v_f^2}{4gs\sqrt{1-\left(\frac{v_i^2+v_f^2}{4gs}\right)^2}}$$With the given values of initial and final speeds and ##s=4.2 ~\mathrm{m}##, I got ##\theta =17^o## and ##\mu_k=0.23## (two significant figures). Needless to say, one has to choose the values of the speeds and the distance traveled so that the ratio giving ##\sin\theta## is less than unity for g = 9.8 m/s².