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 work done by the porter in lowering the suitcase.
(take g = 9.8 ms⁻²)

Solution

Physics sir · 3 steps

The porter pushes up while the suitcase moves down, so his work has to come out negative.

1. Force by porter. The velocity is constant, so the net force is zero. The porter's upward force exactly balances the weight: F = mg = 80 × 9.8 = 784 N, directed up.
2. Direction check. The displacement is 0.80 m down. Force and displacement point opposite ways, so the angle between them is 180° and cos 180° = -1.

   W = Fd cos 180° = 784 × 0.80 × (-1) = -627.2 J

3. So. The work done by the porter is -627.2 J, option (c). Gravity does +627.2 J on the suitcase over the same path, and the two add to zero, as they must at constant speed.

The idea to keep. For work by a force, find the direction of that force and the direction of motion first. Opposite directions mean negative work, whatever the size.

Answer from the NTA final answer key. Solution by PrepWiser.