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What can you say about acceleration if the:

a) static coefficient = kinetic coefficient = 0

b) static coefficient = kinetic coefficient does not equal 0

c) static coefficient is greater than kinetic coefficient

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- Thread starter slayerdeus
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- #1

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What can you say about acceleration if the:

a) static coefficient = kinetic coefficient = 0

b) static coefficient = kinetic coefficient does not equal 0

c) static coefficient is greater than kinetic coefficient

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AKG

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slayerdeus said:

What can you say about acceleration if the:

a) static coefficient = kinetic coefficient = 0

b) static coefficient = kinetic coefficient does not equal 0

c) static coefficient is greater than kinetic coefficient

The first one sounds as if it should be a frictionless surface (ie no friction).

For the answers, are you looking for stuff like, the object is accelerating, deccelerating, or the object is at rest? I'm not sure what kind of answer you are looking for, but it seems to me if the 0 < kinetic = static, I would think the object is accelerating and for the last one, I would think the object would slow down until coming to a rest or something like that.

Does that sound right to anyone else?

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arildno

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slayerdeus said:

What can you say about acceleration if the:

a) static coefficient = kinetic coefficient = 0

b) static coefficient = kinetic coefficient does not equal 0

c) static coefficient is greater than kinetic coefficient

I interpret this question to mean:

Suppose an object is initially at rest on a (horizontal) surface.

What can you say about the acceleration of the object, when a sufficiently strong force to accelerate it is applied to to the object?

a) An arbitrarily small force may be applied in order to accelerate it;

the resulting acceleration will also be arbitrarily small.

b) The sufficient force must satisfy [tex]F>\mu_{s}N[/tex] in order to accelerate the object.

We set [tex]F=\mu_{s}N+\delta{F}[/tex], where [tex]\delta{F}[/tex] can be an arbitrarily small force.

We have, since [tex]\mu_{k}=\mu_{s}[/tex] that the net force on the object, is [tex]\delta{F}[/tex] ; i.e. the resulting acceleration can be made as small as one wish for.

c) The net force is now:

[tex](\mu_{s}-\mu_{k})N+\delta{F}[/tex] that is, the minimum initial acceleration is [tex]a_{min}=(\mu_{s}-\mu_{k})g[/tex]

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