Work & Friction

Question

A person pushes a box of 5kg over a distance of 20m with 50N of force. There is friction acting on the box, which is 20N in the opposite direction to the motion. What is the net work done on the box? Assume that the box starts from rest, what is its final velocity?

Answer 1

1) Net work: W = (50N - 20N)·20m = 600 J

2) By theorem of change of kinetic energy: mv²/2 + W; v = (2W/m)^1/2 = 1200 J/5kg)^1/2 = 15.49 m/s

Answer 2

First, let us find the work done to move the box by 20m.

W = F . d = Fdcos(x) , x is the angle between force and displacement

There are a total of 4 forces acting on the box.

1) Gravity Downwards = -5g N

2) Normal reaction upwards = 5g N

3) Pushing force by the person to the right = 50 N

4) Friction to the left = -20 N

Notice that when calculating the work done the angle x = 90 degrees for gravity and normal reaction.

cos(90) = 0, so W = Fdcos(90) = 0 J

Therefore the work done by gravity and normal reaction is zero.

For friction, angle x = 180 degrees

W = (20)(20)cos180 = 400(-1) = -400 J

For force applied by the person, angle x = 0 degrees

W = (50)(20)cos0 = 1000 J

Net Work done = 1000 - 400 + 0 + 0 = 600 J

To find the final velocity we will use the Work-Energy theorem, which states that

Net Work = Change in K.E = Final K.E - Initial K.E

initial velocity = 0 m/s, so Initial K.E = 0

let final velocity be v m/s, Final K.E = (1/2)5v² J

600 = (5/2)v² - 0

v² = 240

v = 240^1/2 m/s = 15.49 m/s

The net work done on the box is 600 J and its final velocity is 15.49 m/s.