A horizontal 52 Newton force is needed to slide a 50-kg box across a flat surface of a constant velocity of 3.5 m s what is the coefficient of kinetic friction between the box and the floor

μk = 0.106

Explanation:

When the box is moving across the flat surface with a constant velocity, the force applied on it must be equal to the kinetic frictional force applied on the object:

F = μk*W = μk*mg

Where,

F = Force applied on box = 52 N

m = mass of box = 50 kg

g = 9.8 m/s²

μk = coefficient of kinetic friction between box and floor = ?

Therefore,

52 N  = μk(50 kg)(9.8 m/s²)

μk = (52 N)/(490 N)

μk = 0.106

answer: a) y=-8x+5

a) m= -8

b) m= -2

c) m= 7

d) m= 6