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

[EJ25]

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.

 

[cepheid]

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_|| = F_gsin(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_|| = F_f (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_|| > F_f.

 

[EJ25]

what is the right formula to use in this situation?

 

[cepheid]

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 F_f.

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

 

[rwisz]

Your calculations are correct, but there is a small error in your final calculation for the coefficient of friction. You have correctly calculated the magnitude of the frictional force to be 14.7N, but you have used the wrong value for the normal force in your calculation. The normal force should be equal to the component of the weight of the block that is perpendicular to the plane, which is given by Fg*cos30. So the correct equation for the coefficient of friction is:

u = Ff/Fn = 14.7N/29.4*cos30 = 0.58

Therefore, the coefficient of friction between the plane and the block is 0.58. This means that there is a strong frictional force acting between the two surfaces, which is causing the block to slow down as it slides down the inclined plane. If the coefficient of friction were lower, the block would slide down the plane at a faster rate. Conversely, if the coefficient of friction were higher, the block would slide down the plane at a slower rate. The value of the coefficient of friction is dependent on the nature of the surfaces in contact and can vary in different situations.