A) The minimum force required to get the box moving is; F_min = 55.272 N
B) The acceleration of the box if μ_k = 0.33 is; a = 13.96 m/s²
A) We are told that a 12kg box of fruit is to be moved to the other side of the warehouse.
We are given coefficient of static friction; μ_s = 0.47
Formula to get the minimum force is;
F_min = μ_s*mg
F_min = 0.47 * 12 * 9.8
F_min = 55.272 N
B) We are told the force that pushed the box is now constant. This means that;
F_min = μ_k*F_n
where F_n is normal force = ma
a is acceleration
We are given; coefficient of kinetic friction; μ_k = 0.33
Thus;
55.272 = 0.33 * 12a
a = 55.272/(0.33 * 12)
a = 13.96 m/s²
Read more about force with kinetic and static friction at; https://brainly.com/question/14473239
