A person standing on a spring balance inside a stationary lift measures 60 kg. The weight of that person if the lift descends with uniform downward acceleration of 1.8 m/s² will be ______________ N. [g = 10 m/s²]

Solution

Physics sir · 4 steps

In a lift accelerating downward, the floor pushes up less, so the balance reads m(g − a).

1. Mass first. In the stationary lift the balance reads 60 kg, so the person's mass is m = 60 kg. Mass stays the same when the lift moves. Only the reading changes.
2. Pseudo force. In the lift's frame, which accelerates downward at a, add a pseudo force ma pointing upward on the person. The person is at rest in this frame, so N + ma = mg.

   N = m(g - a)

3. Put in numbers. N = 60 × (10 - 1.8) = 60 × 8.2 = 492 N.
4. So. The balance shows the normal force, so the apparent weight is 492 N.

The common slip: 708. Pointing the pseudo force downward gives 60 × 11.8 = 708 N. That is the reading for a lift accelerating upward.

The idea to keep. A balance reads the normal force, not mg. The pseudo force ma always points opposite to the frame's acceleration. Lift accelerating down gives m(g-a). Lift accelerating up gives m(g+a).

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