Trabajo I

Let's see an exercise.
A 40 kg box is dragged 30 m on a horizontal floor, applying a Fp = 100 N exerted by a person. Such force acts doing a 60º angle. The floor exerts a friction force Fr = 20 N. Calculate the work done for each one of these forces Fp, Fr, the weight mg and the normal. Calculate also the net work done on the box.

Solution: There are four forces acting on the box, Fp, Fr, the weight mg and the normal.
The work done by mg and the normal N is zero, because they are perpendicular to displacement(=90º for them).
The work done by Fp is: 
Wp = Fpxcos (using x in place of d) = (100 N)(30 m)cos60º = 1500 J.
The work done by the friction force Fr is: 
Wr = Frxcos180º = (20 N)(30 m)(-1) = -600 J.
The angle between Fr and displacement is 180º because they point in opposite directions.
The net work can be calculated in two equivalents ways:

• As the algebraic sum of the work done by each force:
  W_NET = 1500 J +(- 600 J) = 900 J.
• Finding first the net force on the object along the displacement:
  F_(NET)x= Fpcos - Fr
  and then doing
  W_NET = F_(NET)x = (Fpcos - Fr)x
  
  = (100 Ncos60º - 20 N)(30 m) = 900 J.