By the definition of the work done we have:
a) W=Fdcosθ=120N⋅5.0m⋅cos0∘=600J.W = Fdcos \theta = 120 N \cdot 5.0 m \cdot cos0^{\circ} = 600 J.W=Fdcosθ=120N⋅5.0m⋅cos0∘=600J.
b) W=Fdcosθ=120N⋅5.0m⋅cos30∘=520J.W = Fdcos \theta = 120 N \cdot 5.0 m \cdot cos30^{\circ} = 520 J.W=Fdcosθ=120N⋅5.0m⋅cos30∘=520J.
Answer:
a) W=600J.W = 600 J.W=600J.
b) W=520J.W = 520 J.W=520J.