# 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?

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

Staff Emeritus
Gold Member

## 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 travelling at a constant velocity, suggesting that the net force was zero, suggesting that F|| = Ff (in magnitude).

In this case, the object is NOT travelling 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 travelling down the plane, this suggests the net force is in the "down the plane" direction, requiring F|| > Ff.

EJ25
what is the right formula to use in this situation?

Staff Emeritus