A force of 40.0 N accelerates a 5.0-kg block at 6.0m/s^2 along a horizontal surface. How large is the ....?

Question

A force of 40.0 N accelerates a 5.0-kg block at 6.0m/s^2 along a horizontal surface.

a. how large is the frictional force?
b. what is the coefficient of friction?

Please help me with these questions.
I'm having trouble understanding them.

I have posted this question before, and have seen many different, and varying answers. There seems to be a problem with this question.

Answer 1

Here's the work and answer with g=9.8 m/s^2:
(u=coefficient of friction)

sum of F=ma
40=ma-umg

a) umg is the friction force
umg=ma - 40 = 5(6.0) - 40 = -10 N (aka 10 N going in the opposite direction of the acceleration)

b) u is the coefficient, solve for u
u = (ma - 40) / (mg) = 10 / mg = 10 / 5(9.8) = .20

Answer 2

the friction force is 10.0N
and the coefficent of fricton is 0.2 if acceleration due to gravity is taken as 10m/s^2