Friction of a stone sliding on a pond

In summary, a stone slides across a frozen pond and experiences a deceleration of 8 m/s2. The stone's final speed is 0 because its initial speed is 5 m/s.
  • #1
recon
401
1
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.
 
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  • #2
vf^2 = vi^2 + 2ad

a = -1.6

a = (mu)(g)

mu = .194
 
  • #3
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]
 
  • #4
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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  • #5
Sure

Kinematics:

http://online.cctt.org/physicslab/content/phy1hon/lessonnotes/KinematicsEquations/derivation.asp

Friction:

http://online.cctt.org/physicslab/content/Phy1/lessonnotes/friction/frictionproperties.asp

Newton's Laws:

http://online.cctt.org/physicslab/content/Phy1HON/lessonnotes/forces/Newton4.asp

By the way:

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

It covers 20 meters then it stops so Final speed is 0.

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

[tex] a = \frac{-v^2_{0}}{2\Delta x} [/tex]
 
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  • #6
OK, I have the answer now guys. Thanks a lot.
 

Related to Friction of a stone sliding on a pond

1. What causes the friction between a sliding stone and a pond?

The friction between a sliding stone and a pond is caused by the surface of the stone and the surface of the water. When the stone comes into contact with the water, the molecules of the stone and water interact with each other, creating a force that resists the motion of the stone.

2. Does the friction of a stone sliding on a pond depend on the weight of the stone?

Yes, the friction of a stone sliding on a pond does depend on the weight of the stone. Heavier stones have a greater force pushing down on the water's surface, increasing the friction between the stone and the water.

3. How does the speed of the stone affect the friction?

The speed of the stone does affect the friction. As the stone slides faster, the force of the water molecules pushing against it increases, resulting in greater friction between the stone and the water.

4. Can the surface tension of the pond affect the friction of a sliding stone?

Yes, the surface tension of the pond can affect the friction of a sliding stone. Surface tension is the cohesive force between molecules at the surface of a liquid, and it can create resistance against the motion of objects on the surface of the water, including a sliding stone.

5. How does the shape of the stone impact the amount of friction on a pond?

The shape of the stone can impact the amount of friction on a pond. A flat, smooth stone will have less friction compared to a rough, irregularly shaped stone due to the differences in surface area and the contact between the stone and the water molecules.

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