A 20 kg block slides across a horizontal surface if the coefficient of kinetic friction between the block and the surface is 0.20, what is the force of kinetic friction that acts on the block

I got:

u

s

=

0.51

F

=

78.4

N

Explanation:

At the start the  

100

N

force is just enough to overcome static friction so we can write:

Force

=

Static Friction

F

=

μ

s

N

where  

μ

s

is the coefficient of static friction and  

N

=

Normal Reaction that in an horizontal case such this will be equal to the weight of the block, so  

N

=

m

g

.

We get:

F

=

μ

s

⋅

m

g

in numbers:

100

=

μ

s

⋅

20

⋅

9.8

μ

s

=

100

20

⋅

9.8

=

0.51