In the figure above the work done by the force ... - JAMB Physics 2005 Question

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