Calculating Distance with Friction on an Icy Surface

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The discussion focuses on calculating the friction coefficient and distance traveled by a disc moving on an icy surface. The correct friction coefficient is determined to be 0.17, derived from the equation μ = Vo²/2gx. For a second disc with double the mass, the distance traveled remains the same due to the cancellation of mass in the friction force calculation. The conversation highlights the importance of not substituting values too early in problem-solving to avoid unnecessary complications. Overall, the relationship between mass and distance traveled is clarified, emphasizing that increased mass does not lead to reduced distance on an icy surface.
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Homework Statement


Hi, would like to get a feedback on my answer to this question, did I do it right?
In part 2 of the question is there another way to calculate distance with friction?

the question is:
a disc is moving on icy surface has an initial speed of 12m/s 42m until it stops
1. what is the friction coefficient?
2. if you take a disc made of the same material as the first and moves at the same initial speed but has double the mass how far would that disc go?

ty!

Homework Equations


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The Attempt at a Solution



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You are correct. However you substituted values too early which made things a bit harder than necessary. If you hadn't done that then for Q1 you would arrive at the equation...

μ = Vo2/2gx

then substitute values to give μ = 0.17

For Q2 you can simply rearrange the equation to give...

x = Vo2/2gμ

from which you can see that x is independent of the mass.
 
PS: The more mass the disc has the more inertia it has so you might expect it to go further BUT the friction force is proportional to the normal force which depends on mass. So mass ends up cancelling.

PPS: It's not unknown for examiners to give you lots of unnecessary information and if you substitute values early you can end up wasting a lot of time (and accuracy) if it cancels out of the equation.
 
tyvm!
CWatters said:
PS: The more mass the disc has the more inertia it has so you might expect it to go further
yes, I wondered about that but the other way around, thought it will travel less, it seemed to make sense intuitively - that heavier objects will travel less distance, obviously it doesn't - why? is that due to inertia as you mentioned?
CWatters said:
It's not unknown for examiners to give you lots of unnecessary information and if you substitute values early you can end up wasting a lot of time (and accuracy) if it cancels out of the equation.
ty!, noted!
 
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Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
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