What will be the weight of a man of mass 60kg standing in a lift if the lift is descending vertically at 3ms$^2$?

Analyze the question: A 60 kg man stands in a lift that accelerates downward at 3 m/s² (the unit should be m/s², not ms²). We need the force the floor exerts on him, which is his apparent weight.

Concept: When a lift accelerates downward, the apparent weight (the normal reaction R) is less than the true weight:

mg − R = ma, so R = m(g − a)

Solution:

Taking g = 10 m/s²:

R = 60(10 − 3) = 60 × 7 = 420 N

Taking g = 9.8 m/s²:

R = 60(9.8 − 3) = 60 × 6.8 = 408 N

The closest option is 420 N, which uses g = 10 m/s², the usual JAMB value.

Checking the other options:

• 660 N = 60(10 + 3) = 780 N? No, 60 × 11 = 660 would need a = 1 upward, so it does not fit.
• 780 N = 60(10 + 3), which is what you would get if the lift were going up with acceleration 3 m/s².
• 546 N is not obtained by any correct use of the formula.

Answer: The apparent weight is 420 N, which is option D.

Tip: Remember the direction rule:

• Lift accelerating up → R = m(g + a), and you feel heavier.
• Lift accelerating down → R = m(g − a), and you feel lighter.

Also, watch the wording "descending at 3 m/s²". If it had said "constant velocity", then a = 0 and the weight would just be 600 N.