What's the coefficient of kinetic friction?

[Austin Gibson]

Homework Statement

:

A sled starts from rest at the top of a snow-covered incline that makes a 20° angle with the horizontal. After sliding 68 m down the slope, its speed is 17 m/s. Use the work–energy theorem to calculate the coefficient of kinetic friction between the runners of the sled and the snowy surface.
[/B]

Homework Equations

:

1. Fcos(theta)*r = work
2. .5*m*(v(final)^2 - v(initial)^2) = delta kinetic energy[/B]

The Attempt at a Solution

: (IGNORE THE NEGATIVE SIGN AT THE END)
[/B]

 

[kuruman]

The coefficient of kinetic friction must be positive. Check your arithmetic.

 

[haruspex]

The coefficient of kinetic friction must be positive. Check your arithmetic.

The OP says to ignore the minus sign, presumably because Austin noticed, after taking the image, that it had been erroneously introduced in the final step.
The (positive) answer looks right to me.

 

[kuruman]

The OP says to ignore the minus sign, presumably because Austin noticed, after taking the image, that it had been erroneously introduced in the final step.
The (positive) answer looks right to me.

Right. I started responding when there was no attempt at a solution and my original response referred to that. Then my screen went blank and I was kicked out of the thread. When I reentered, I saw the image with the attempt at a solution but did not see OP's caution to ignore the negative sign. Not that it matters.

 

[Austin Gibson]

I submitted "0.13" as my answer. That answer was rejected. Proof:

 

[Austin Gibson]

The OP says to ignore the minus sign, presumably because Austin noticed, after taking the image, that it had been erroneously introduced in the final step.
The (positive) answer looks right to me.

I submitted "0.13" as my answer, but when I copied my calculations from paper to the whiteboard to check it from a broader perspective and capture a picture, I forgot that the negatives canceled when you divide (144.5-228)/626 at the end. Nonetheless, I'm still stuck.

 

[haruspex]

I submitted "0.13" as my answer, but when I copied my calculations from paper to the whiteboard to check it from a broader perspective and capture a picture, I forgot that the negatives canceled when you divide (144.5-228)/626 at the end. Nonetheless, I'm still stuck.

You are saying .13 is rejected?
I get .133, so maybe it wants another digit, but it should not require that since the data given are only to two significant figures.

 

[Austin Gibson]

You are saying .13 is rejected?
I get .133, so maybe it wants another digit, but it should not require that since the data given are only to two significant figures.

Your instincts were correct, but that's ridiculous because, as you mentioned earlier, the measurements are listed in TWO significant figures... THANK YOU!

 

[kuruman]

You are saying .13 is rejected?
I get .133, so maybe it wants another digit, but it should not require that since the data given are only to two significant figures.

Your instincts were correct, but that's ridiculous because, as you mentioned earlier, the measurements are listed in TWO significant figures... THANK YOU!

These machine-graded problems have a global accuracy tolerance for deciding whether the given answer is correct. So the course instructor is responsible for making sure that the number of significant figures in the given quantities is consistent with the tolerance. I advise @Austin Gibson to ask the course instructor what the tolerance is for future reference and to use at least 3 significant figures for machine-graded answers.