JAMB - Physics (2005 - No. 37)

In the figure above, the work done by the force of 100N inclined at an angle of 60^o to the object dragged horizontally to a distance of 8m is

800J

600J

100J

400J

Explanation

Since the force is inclined at an angle 60^o to the horizontal, the effective component of te force along the horizontal is given by F x cos 60^o
= 100 x 0.5 = 50N.
∴ work done = F x cos 60 x displacement
= 50 x 8.0m
= 400.0J

JAMB - Physics (2005 - No. 37)

Explanation

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