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JAMB - Physics (2005 - No. 37)

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

Penjelasan

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

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