A box of mass 12 kg sits on a level surface. The coefficient of friction between the box and the floor is 0.25

Question

A box of mass 12 kg sits on a level surface. The coefficient of friction between the box and the floor is 0.25. A rope attached to the box pulls it horizontally to the right with a force of 38N. What are all the forces acting on the box?

a) what is the magnitudes of the friction force?

b) what is the acceleration of the box?

c) if the rope maintains this same force for 3 seconds and the box starts from rest, how far does the box travel?

Answer 1

Ok, let's start with the forces :

The normal force = 12g

the weight = 12g

the force (F) = 38 N

the friction force = (Normal force)*u

u = 0.25, then :

normal force = 117.6 N

force of friction = 29.4 N

a) the force of friction is f = 29.4 N

b) Here we have to use the second law of Newton.

F - f = Ma

a = acceleration

38 - 29.4 = 12a

a = 0.72 m/s^2

c) the box starts from rest, then initial speed = 0

acceleration = 0.72 and time = 3 seconds

we can use this formule : x = Vo*t + a*t^2 / 2

where Vo = 0 and x = distance

x = 0.72*9 / 2 = 3.24 meters

Hope that might help you