For this case we have that by definition, the mechanical power is given by:
[tex]P = \frac {W} {t}[/tex]
Where:
W: It is the work done
t: It's time
For its part:
[tex]W = F * d[/tex]
Where:
F: It is the applied force
d: It is the distance traveled
According to the data we have that the drag force of the tractor is [tex]F = 300lb[/tex], while the distance traveled is [tex]d = 80ft.[/tex]
Substituting:
[tex]P = \frac {300 * 80} {120}\\P = \frac {24000} {120}\\P = 200[/tex]
Finally, the power is:
[tex]\frac {200lb-ft} {s}[/tex]
Answer:
Option C
