A porter lifts a heavy suitcase of mass 80 kg and at the destination lowers it down by a distance of 80 cm with a constant velocity. Calculate the workdone by the porter in lowering the suitcase. (take g = 9.8 ms^–2)

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+627.2 J

–62720.0 J

–627.2 J

784.0 J

answer is D.

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Detailed Solution

Because the luggage is being lowered at a steady speed, there is no net force.

Net force = 0

$F_{ext} = mg$

$Work done = F_{ext} scos 180^{\circ}$

$= mg \times s \times - 1 = 80 \times 9.8 \times 0.8 \times - 1 = - 627.2 J$

Hence the correct answer is -627.2J.

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