Asked by Grant

A force of 23.0 N is required to start a 3.1 kg box moving across a horizontal concrete floor.
(a) What is the coefficient of static friction between the box and the floor?

(b) If the 23.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?



How do I find the answer to part b?
Answered by bobpursley
applied force-friction=mass*acceleration
solve for friction, then set it equal to mu*mg and solve for mu.
Answered by Grant
so i would do.. 23N - friction= 3.1(or 30.38N) *.5

and get f=7.81

then 7.81 = mu k * mu g ?
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