A man carries a bucket of weight 60 N and wa | vedclass

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Question

(A) The total work done by the man is the sum of work done in the horizontal and vertical directions.Work done in the horizontal direction $(W_{Hz})$:

Vedclass · Accepted Answer

$300$

Vedclass · Answer

(A) The total work done by the man is the sum of work done in the horizontal and vertical directions.Work done in the horizontal direction $(W_{Hz})$: Since the force (weight of the bucket) acts vertically downwards and the displacement is horizontal,the angle $	heta = 90^{\circ}$. Thus,$W_{Hz} = F \cdot d \cdot \cos(90^{\circ}) = 0 \ J$.Work done in the vertical direction $(W_{vert})$: The man lifts the bucket against gravity,so the force applied is equal to the weight $(60 \ N)$ and the displacement is $5 \ m$ upwards. Thus,$W_{vert} = F \cdot d \cdot \cos(0^{\circ}) = 60 \ N 	imes 5 \ m = 300 \ J$.Total work done $W_{net} = W_{Hz} + W_{vert} = 0 \ J + 300 \ J = 300 \ J$.

$A$ man carries a bucket of weight $60 \ N$ and walks $7 \ m$ in the horizontal direction,then climbs $5 \ m$ in the vertical direction. What is the work done by the man in $J$?

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