If a 68kg object is being pulled across the hard floor with a horizontal force of 400N...(force problem)

1. If a 68kg object is being pulled across the hard floor with a horizontal force of 400N...

a. What would the acceleration of the object be if it was on a frictionless floor?
b. if there were friction, what would the force of friction be if it was pulled at a constant velocity?
c. If it wasn't at a constant velocity, but instead accelerating at 3.2m/s^2, what would the force of friction be acting on the object?
d. What would the coefficient be when the object accelerates at 3.2m/s^2

Thanks!

m = 68 kg
F = 400 N

a) The acceleration of the object : a = F/m = 400/68 = 5.88 m/s^2
b) The force of friction, must be equal with F if it was pulled at a constant velocity ->friction f =F= 400 N
c) if the acceleration at 3.2 m/s^2 : ΣF = m*a
F - fk = m * a
400 - fk = 68 * 3.2 ----> fk = 400 - 217.6 = 182.4 N
d) fk = µ * N = µ * m * g
182.4 = µ * 680 ---> µ = 182.4/680 = 0.27

mass = 68 kg,
Force, F = 400 N
a)
Acceleration = 400/68 = 6.88 m/s^2 >===========< ANSWER

b)
At constant velocity,
Force of Friction = -400 N >===============< ANSWER

c)
Acceleration = 3.2 m/s^2
Force = 68*3.2 = 217.6 N

Force of Friction = 400 -217.6 = 182.4 N >=======< ANSWER

d)
Coefficient, µ = 182.4/400 = 0.456 >===========< ANSWER