Friction of a stone sliding on a pond

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Homework Help Overview

The problem involves a stone sliding across a frozen pond, with given initial speed and distance before coming to rest. The goal is to calculate the coefficient of friction between the stone and the pond's surface, exploring concepts related to kinematics and friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between deceleration, frictional force, and mass. There are attempts to apply kinematic equations and Newton's laws to derive the coefficient of friction. Some participants express confusion regarding the variables and calculations used.

Discussion Status

The discussion includes various approaches to the problem, with some participants offering links to external resources for further clarification. There is acknowledgment of mistakes in calculations, and while one participant claims to have found the answer, the overall conversation reflects a mix of interpretations and ongoing exploration.

Contextual Notes

Participants note confusion regarding the definitions of variables and the relationships between forces. There is also mention of the algebra becoming complicated, indicating potential challenges in the problem setup.

recon
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This isn't really homework because I'm not doing it for school. I'm on my holidays at the moment, and I'm trying to do a bit of Physics on my own. As you can imagine, I have not gotten far, and am stuck on this problem:

A stone slides in a straight line across a frozen pond. Given that the initial speed of the stone is 5m/s and that it slides 20m before coming to rest, calculate the coefficient of friction between the stone and the surface of the frozen pond.

The deceleration of the stone is (20*2)/5 = 8 m/s2.

How should I proceed next?

EDIT: I'm really confused. Is the resultant force = 8 X mass?

Is the resultant force = frictional force - forward force?

I know that frictional coefficient = Frictional force/contact force.

However, I've done the algebra and it comes out very messy.
 
Last edited:
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vf^2 = vi^2 + 2ad

a = -1.6

a = (mu)(g)

mu = .194
 
Use the relation

[tex]v^2 = v^2_{0} + 2a \Delta x[/tex]

and the fact that the only force acting on the stone with a nonzero horizontal component is friction.

Newton's 2nd Law

[tex]\mu mg = ma[/tex]
 
I don't really understand the variables that the two of you have used above. Is there a site that you can refer me to for this?

OK, I see my careless mistake. The deceleration is not 8 m/s2, that is the time taken for it to decelerate. So the deceleration is 5/8 = 0.625 m/s2, is it not?
 
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OK, I have the answer now guys. Thanks a lot.
 

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