Calculate the apparent weight loss of a man weighing 70kg in an elevator moving downwards with an acceleration of 1.5$ms^{-2}$. [g = 10$ms^{-2}$]

Analyze the question: A man of mass 70 kg stands in an elevator accelerating downwards at 1.5 m/s². We want the apparent weight loss, which is the difference between his true weight and the reading on a scale (the normal reaction).

Concept:

• True weight: W = mg
• Elevator accelerating downwards: the net force downward is mg − R = ma, so R = m(g − a)
• Apparent weight loss = W − R = ma

Solution:

1. True weight: W = mg = 70 × 10 = 700 N
2. Apparent weight: R = m(g − a) = 70 × (10 − 1.5) = 70 × 8.5 = 595 N
3. Apparent weight loss = 700 − 595 = 105 N

Check: ma = 70 × 1.5 = 105 N ✓

Answer: The apparent weight loss is 105 N, which is option D.

Note that 595 N (option B) is the apparent weight, not the loss. This is the trap in the question, so read what is being asked carefully.

Exam tip: For "apparent weight loss" or "gain", the answer is always just ma.

• Elevator going down with acceleration: apparent weight = m(g − a), loss = ma
• Elevator going up with acceleration: apparent weight = m(g + a), gain = ma