A force of 40.0 N is required to start a 5.0-kg box moving across a horizontal concrete floor...?

Question

a) What is the coefficient of static friction between the box and the floor?
b)If the 40.0 N force continues, the box accelerates at 0.70 m/s^2. What is the coefficient of kinetic friction?

Answer 1

I will assume the force is applied horizontally.
Then
40=5*g*u for static

after moving, kinetic will follow
F=m*a
so
40-5*g*u=5*0.70

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