Physics-friction and coefficient of kinetic friction question.?

Vertical forces:

Normal force: N (upward)

Weight: W = 300 N (downward)

Thus, N = W = 300 N

Horizontal Forces:

Applied force: F = 50 N (forward)

Frictional force: Ff = u*N (backward)

Thus, F = Ff = u*N

So, u = F/N = (50 N)/(300 N) = 0.167

If there was no friction, then the net force is just the applied force:

F(net) = 50 N

And, the mass of the block is:

m = W/g = (300 N)/(9.8 m/s^2) = 30.6 kg

So, the acceleration is:

a = F(net)/m = (50 N)/(30.6 kg) = 1.63 m/s^2

u=Fk/Fn

Because it is moved at a constant speed I believe your Fk is 50 N.

And you are right about the normal force being 300N

I think you would just do 50N/300N=.167

This answer makes sense because the mu value is less than one.

I understand how you feel about physics.