forces help?

Question

A force of 27.0 N is required to start a 3.9 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 27.0 N force continues, the box accelerates at 0.50 m/s2. What is the coefficient of kinetic friction?

Answer 1

f=uN
a)for static case
u= f/mg
u=27.0 / 3.9 x 9.81=
u=.706

b) For kinetic case
u= f/mg where f= F-ma
u= (F-ma)/mg
u= (27.0 - 3.9 x 0.5)/ 3.9 x 9.81=
u= 0.655

Answer 2

in the first, if its not accelerating yet, then the sum of the forces in the x direction is zero, so:
F - friction = 0 (F is 27N and friction is negative since opposite)
so F = friction but u know that friction = coeff*N
F = coeff*N you also know that N = mg (sum of all foces in the y direction = 0)
so F = coeff*m*g
coeff = F/(mg)

for the second one, the sum of the forces is no longer zero, but rather m*a
F - friction = ma
F - coeff*m*g = m*a
coeff = (F-ma)/(mg)

the difference between the two coefficients is that the first is the one while the box is at rest (static), and the other is the one while its in motion (kinetic), and thats y they are different.