A force of 40N continues to accellerate a 5kg box on a flat surface. If the box accellerates at 0.7m/s^2, whats the coefficent of kinetic f?

The net force in this scenario would be the applied force of 40 N minus the force of friction. Because the net force is also mass times acceleration according to Newton's Second Law, we can set the two equal to each other.

Fnet = F - Ff

Fnet = ma

F - Ff = ma

Solve for Ff:

Ff = F - ma

Now we need to solve for the coefficient of kinetic friction μ. It is defined by the following equation:

Ff = μFn (Fn is the normal force)

Now we need to find Fn:

Fn = Fgrav

Fn = mg (g = 9.81 m/s^2)

Substitutions:

Ff = μmg

μmg = F - ma

μ = (F - ma)/(mg)

Now substitute in values to find your answer:

μ = (40 N - 5 kg * 0.7 m/s^2)/(5 kg * 9.81 m/s^2) = 0.744 (This number can be rounded depending on number of sig figs given in problem.)