What is the coefficient of friction between the plane and the block?

AI Thread Summary
A 3 kg block slides down a 30-degree incline with a constant acceleration of 0.5 m/s², prompting the question of the coefficient of friction between the block and the plane. The normal force (Fn) was calculated as 25.46 N, and the frictional force (Ff) was found to be 14.7 N. However, the discussion highlights a misunderstanding regarding the forces acting on the block, emphasizing that since the block is accelerating, the net force down the incline must be greater than the frictional force. The correct approach involves applying Newton's second law to account for the net force, which combines both the gravitational component and friction. Understanding these principles is crucial for solving the problem accurately.
EJ25
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Homework Statement


A 3 kg block slides down a 30 degree inclined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2 m. What is the coefficient of friction between the plane and the block?


Homework Equations


Ff=uFn
Fn=Fg*cosA
Ff=Fg*sinA



The Attempt at a Solution


Fn=(3)(9.8)*cos30
Fn=29.4*cos30
Fn=25.46N

Ff=(3)(9.8)*sin30
Ff=29.4*sin30
Ff=14.7N

Ff=uFn
14.7/25.46=u
u=0.58

i am doing something wrong here and i don't know what it is please help.
 
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EJ25 said:

Homework Statement


A 3 kg block slides down a 30 degree inclined plane with constant acceleration of 0.5 m/s2. The block starts from rest at the top. The length of the incline is 2 m. What is the coefficient of friction between the plane and the block?


Homework Equations


Ff=uFn
Fn=Fg*cosA
Ff=Fg*sinA

This part in red is NOT TRUE in this situation! F|| = Fgsin(A) is the component of the weight that acts parallel to the plane, trying to pull the object down it. You would only equate that to the frictional force (in magnitude) IF the object were traveling at a constant velocity, suggesting that the net force was zero, suggesting that F|| = Ff (in magnitude).

In this case, the object is NOT traveling at a constant velocity. It is accelerating down the plane, suggesting the net force parallel to the plane is NOT zero. In fact, since the object is traveling down the plane, this suggests the net force is in the "down the plane" direction, requiring F|| > Ff.
 
what is the right formula to use in this situation?
 
EJ25 said:
what is the right formula to use in this situation?

Physics is not about memorizing formulae; it is about understanding and applying physical principles. Since you know the acceleration, you know the NET force in the || direction.* You also know that the net force has to be equal to the sum of the forces acting along that direction. There are are two such forces. One is F|| and the other is Ff.

*This knowledge comes as a result of Newton's second law, which is the principle that is applicable here.
 

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