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Q.

A force of 100N is acting on a 10kg block at  60∘ above the horizontal. If it travels 5m in 10s work done by force is

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a

500J

b

250J

c

2500J

d

150J

answer is B.

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Detailed Solution

work done w=Fscosθ work done=100×5×cos60=100×5×12 Work done=250Jhere F=force; S=displacement; θ=angle between force and displacement
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