A 50-kg box is being pushed a distance of 8.0 m across the floor by a force P with arrow whose magnitude is 159 N. The force P with arrow is parallel to the displacement of the box. The coefficient of kinetic friction is 0.24. Determine the work done on the box by each of the four forces that act on the box. Be sure to include the proper plus or minus sign for the work done by each force. (The floor is level and horizontal. Forces that act on the box: pushing force P, frictional force f, normal force N, and weight mg.)

Applying the definition of work, as the product of the component of the force applied, along the direction of the displacement, times the displacement, we find that due to the normal force is always perpendicular to the surface, it does no work, as it has not a component in the direction of the displacement, so Wn = 0.
As the weight goes directly downward, it has no component in the direction of the displacement either, so Wg = 0.
We can get the work done by the force applied P, simply as follows:

[tex]W_{p} = F_{p} * d * cos \theta = 159 N * 8.0 m * cos 0 = 1,272 J (1)[/tex]

Finally, we have the work done by the force of friction that always opposes to the displacement, so it has negative sign.
The frictional force , can be written as follows:

[tex]F_{fr} = \mu k * F_{n} (2)[/tex]

where μk = coefficient of kinetic friction = 0.24

Fn = Normal Force

In this case, since the surface is level and horizontal, and there is no acceleration of the box in the vertical direction, this means that the normal force (in magnitude) must be equal to the weight:
Fn = m*g = 50 kg * 9.8 m/s2 = 490 N
Replacing these values in (2), we get:

[tex]F_{fr} = 0.24 * 50 kg* 9.8 m/s2 = 117.6 N[/tex]

Applying the definition of work, we can get the work done by the frictional force, as follows:

[tex]W_{ffr} = F_{fr} * d * cos \theta = F_{fr} * d * cos (180) = - F_{fr} * d = -117.6 N * 8.0 m \\ = W_{ffr} = -940.8 N[/tex]