In the figure above, the work done by the force of 100N inclined at an angle of 60o to the object dragged horizontally to a distance of 8m is
- 800J
- 600J
- 100J
-
400J
The correct answer is: D
Explanation
Since the force is inclined at an angle 60o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60o= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J