Question

A block of metal of mass 0.05 kg when placed over a smooth plane inclined at 15 slides down without acceleration the inclination is now increased by 15" What should be the acceleration of the block

Answer 1

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

Notice that I have defined a rotated set of axes and I labelled them x’ and y’. The x’-axis is parallel to the plane and the y’-axis is perpendicular to the plane. I chose positive x’-axis down the plane since the block is accelerating down the plane. Now write Newton’s second law in the x’ direction:

ΣFx′=max′

The component of the weight (mg) acting down the plane is found by resolving the weight into components as shown below:

So the component of the weight acting down the plane is (mg)sin30. The friction force acts opposite the direction of motion (up the plane) as shown on my free body diagram.

ΣFx′=max′

(mg)sin30−Ffric=max′

(0.5)(9.81)sin30−Ffric=(0.5)(3ms2)

Ffric=0.953N

Once you know the friction force, you can determine the coefficient of friction using:

Ffric=μkFN

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

ΣFy′=may′

Since the block is not “lifting” off the ramp, there is no motion in the y’-direction, so ay′=0.

ΣFy′=0

FN−(mg)cos30=0

FN=(0.5)(9.81)cos30=4.25N

Ffric=μkFN

0.953N=μk(4.25N)

μk=0.224