Question

a block slides down on an inclined at 30° with acceleration equal to g/4. find the coefficient if kinetic friction​

Answer 1

First draw a free body diagram of the block. A free body diagram shows all the forces acting on the object.

a rotated set of axes - x’ and y’. The x’-axis is parallel to the inclined plane and the y’-axis is perpendicular to the plane. I chose positive x’-axis down the plane. The component of the weight (mg) acting down the plane is found by resolving the weight into components as shown below using θ=30 :

So the component of the weight acting down the plane is (mg)sinθ. The friction force acts opposite the direction of motion (up the plane) as shown on my free body diagram. To determine Ffric , write Newton’s second law in the x’ direction:

ΣFx′=max′

(mg)sinθ−Ffric=max′

but the block is moving at constant speed, so ax′=0

∴ (mg)sinθ−Ffric=0

or

Ffric=(mg)sinθ ————— equation 1

Since the block is sliding, we have the maximum possible friction force given by:

Ffric=μkFN ————— equation 2

To determine FN , write Newton’s law in y’ direction:

ΣFy′=may′

The block is not lifting off the incline, so ay′=0

∴ FN−(mg)cosθ=0

FN=(mg)cosθ

substitute this into equation 2:

Ffric=μk(mg)cosθ ————— equation 3

combine equations 1 and 3:

(mg)sinθ=μk(mg)cosθ

or

μk=sinθcosθ=tanθ

μk=tan30=0.577

Answer 2

Answer:

F

Explanation: