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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

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