Answer:
A. 780 J
B. 120 J
C. 660 J
Explanation:
From the question given above the following data were obtained:
Dragging force (Fₔ) = 65 N
Distance (s) = 12 m
Force of friction (Fբ) = 10 N
A. Determination of the work done by the dragging force.
Dragging force (Fₔ) = 65 N
Distance (s) = 12 m
Workdone (Wd) by dragging force =?
Wd = Fₔ × s
Wd = 65 × 12
Wd = 780 J
Therefore, the work done by the dragging force is 780 J
B. Determination of the work done by friction.
Distance (s) = 12 m
Force of friction (Fբ) = 10 N
Workdone (Wd) by friction =?
Wd = Fբ × s
Wd = 10 × 12
Wd = 120 J
Therefore, the work done by friction is 120 J
C. Determination of the net work done on the box.
Dragging force (Fₔ) = 65 N
Distance (s) = 12 m
Force of friction (Fբ) = 10 N
Net work done (Wd) =?
Next, we shall determine the net force acting on the box. This can be obtained as follow:
Dragging force (Fₔ) = 65 N
Force of friction (Fբ) = 10 N
Net force (Fₙ) =?
Fₙ = Fₔ – Fբ
Fₙ = 65 – 10
Fₙ = 55 N
Thus, the net force acting on the box is 55 N
Finally, we shall determine the net work done on the box as follow:
Distance (s) = 12 m
Net force (Fₙ) = 55 N
Net work done (Wd) =?
Wd = Fₙ × s
Wd = 55 × 12
Wd = 660 J
Therefore, the net work done on the box is 660 J
