A person pushes horizontally on a heavy box and slides it across the level floor at constant velocity. The person pushes with a 60.0 N force for the first 6.88 m, at which time he begins to tire. The force he exerts then starts to decrease linearly from 60.0 N to 0.00 N across the remaining 6.88 m. How much total work did the person do on the box?

Then, the force is variable, so, we should calculate the work done at every instant during the 6.88 m and add all that work. Mathematically, that is expressed as follows:

[tex]W = \int\limits^x_0 {F} \, dx[/tex]

Where F is the force and dx is an infinitesimally small traveled distance.

The applied force F depends on the traveled distance and decrease linearly. Initially, the applied force is 60.0 N, i.e., when the traveled distance is zero, the force is 60.0 N. The function F(x) (force as a function of traveled distance) could be written as follows:

F(x) = 60 - mx

Where x is the distance and m is the rate of decrease of the force (the slope of the function). The slope is calculated as follows:

m = ΔF / Δx

Where:

ΔF = change in the applied force (final force - initial force)

Δx = change in the traveled distance ( final position - initial position)

Then:

m = -60.0 N / 6.88 m

The function F(x) will be:

F(x) = 60 - (60/6.88)x

The work done will be:

[tex]W = \int\limits^x_0 {(60 - (60/6.88 )x)} \, dx[/tex]

Integrating from 0 to x = 6.88 we obtain:

W = [60 x - (30/6.88) x² ] from x = 0 to x = 6.88

W = 60 (6.88) - (30/6.88) (6.88)² = 206 J

The total work done by the person is:

W = 413 J + 206 J = 619 J