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