Coefficient of static & kinetic friction

AI Thread Summary
To find the coefficient of static friction for a 2.5 kg object requiring 4.90 N to start moving, the calculation uses the formula u = f/N. Substituting the values, u is calculated as 0.2, which represents the coefficient of static friction. The discussion highlights confusion regarding the distinction between static and kinetic friction, emphasizing that static friction must be overcome to initiate motion. Clarification is sought on whether the calculated value pertains to static friction, given the context of the problem. The solution provided is indeed correct for determining the coefficient of static friction.
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Homework Statement


If the force needed to start the motion of an object having a mass of 2.5 kg is 4.90 N, what is the coefficient of static friction?

Homework Equations


f = uN
Since u is missing, the formula is u = f/N

The Attempt at a Solution


u = 4.90 N/(2.5 kg X 9.8 m/s2)
u = 4.90 N/24.5 N
u = 0.2

The problem is asking for the coefficient of static friction. I'm not sure if I got the answer for kinetic friction? I know the difference between the two but I don't know why the problem required static friction when it is talking about a moving object.
 
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When an object is at rest it will have static friction and kinetic friction when it is moving. So in order to move an object, you will need to overcome the static friction.
 
semc said:
When an object is at rest it will have static friction and kinetic friction when it is moving. So in order to move an object, you will need to overcome the static friction.

Thanks semc. But is my solution correct?
 
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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