A 4.10 kg block is pushed along a floor by a constant applied force that is horizontal and has a magnitude of?

Question

A 4.10 kg block is pushed along a floor by a constant applied force that is horizontal and has a magnitude of 37.0 N. Figure 6-26 gives the block's speed v versus time t as the block moves along an x axis on the floor. What is the coefficient of kinetic friction between the block and the floor?

for the table
the speed is 5m/s.

Answer 1

Since it is moving with a constant speed (who cares what number it is so long it is not tooooo fast)

F=f=uN
N=mg

so
u= F/mg= 37.0/(4.10 x 9.81) =
u=.92

Answer 2

Ff = friction stress Fn = regular stress ok = coefficient of kinetic friction because of the fact it is going horizontal then Fn = 4.40 kg x 9.8 m/s² = 40 3.12 N Ff = ok(Fn) 40 two.0 = ok(40 3.12) ok = 40 two.0/40 3.12 = 0.ninety seven