Answer:
181.82N
Explanation:
We have the following:
Coefficient of kinetic friction: [tex]\mu=0.22[/tex]
Kinetic frictional force: [tex]f=40N[/tex]
and we have a formula that relates these two quantities:
[tex]f=\mu N[/tex]
where [tex]N[/tex] is the normal force. So since the normal force is what we are asked for, we clear for N in the last equation:
[tex]N=\frac{f}{\mu}[/tex]
and substituting all the known values:
[tex]N=\frac{40N}{0.22} =181.818N[/tex]
The normal force has a value of 181.82N
