A 4 kg block is placed on an incline at an angle of 30 degrees. If the coefficient of static friction is 0.3 and the coefficient of kinetic friction is 0.2 what is the net force on the block?

Answer 1

I hope someone validates this answer since I don't play around with these concepts all of the time, but here is my response.

The force that the block exerts down the ramp is the following:

Fof block=mg*Sin(30)

Fof block=(4.0kg)*(9.8m/s^2)*(0.5)=19.6N

Force of Static Friction is the following:

Fsk=Ms*FN=(0.9)[mg*Cos(30)]

Fsk=(4.0kg)*(9.8m/s^2)*(0.9)=35.3N

Fsk>Fof Block, therefore the block doesn't go down the slide.

The force exerted by kinetic friction is only relevant if the force of the block is able to overcome the force exerted by static friction, therefore the force exerted on the block is equal to what it exerts on static friction from Newton's third law.

So, the answer is

19.6N, the force that block exerts down the incline.

The force of kinetic friction is irrelevant and is not needed for this problem.