Aforce of 36.0 n is required to start a 3.0-kg box moving across a horizontal concrete floor. part a) what is the coefficient of static friction between the box and the floor? express your answer using two significant figures. part b) if the 36.0-n force continues, the box accelerates at 0.54 m/s^2. what is the coefficient of kinetic friction? express your answer using two significant figures.

A) 1.2

B) 1.1

Explanation:

A)

F = force required to start the box moving = 36.0 N

m = mass of the box = 3 kg

 = weight of the box = mg = 3 x 9.8 = 29.4 N

 = normal force acting on the box by the floor

normal force acting on the box by the floor is given as

 = = 29.4

 = Static frictional force = F = 36.0 N

= Coefficient of static friction

Static frictional force is given as

 =

36.0  = (29.4)

= 1.2

B)

a = acceleration of the box = 0.54 m/s²

F = force applied = 36.0 N

= kinetic frictional force

= Coefficient of kinetic friction

force equation for the motion of the box is given as

F - = ma

36.0 - = (3) (0.54)

= 34.38 N

Coefficient of kinetic friction is given as

= 1.1

what is the question?

the answer is a 3.8cm

explanation: