Elevator Dynamics – Apparent Weight on a Scale During Descending Motion A man stands on a spring scale in a lift. At rest, the scale reads 60 kg. When the lift begins to descend with constant acceleration equal to g/5, what reading will the scale indicate (in kilograms)?

Introduction / Context:
Apparent weight measured by a scale is the normal reaction between the person and the scale, not the true weight. In accelerating frames like elevators, Newton’s second law modifies the normal reaction depending on the direction and magnitude of the elevator’s acceleration.

Given Data / Assumptions:

• Scale reading at rest = 60 kg → mass m = 60 kg (since scale is calibrated in kg as m = W/g).
• Lift accelerates downward with a = g/5.
• Downward is taken positive for acceleration; neglect cable mass and friction.

Concept / Approach:

For a person of mass m in a descending elevator, the normal reaction N equals m(g − a). A scale calibrated in kilograms effectively displays m_eff = N/g. Therefore, the indicated mass reduces by the factor (1 − a/g) for downward acceleration.

Step-by-Step Solution:

Verification / Alternative check:

Limiting cases: if a = 0 → reading 60 kg (matches rest). If a → g (free fall) → reading 0 kg, consistent with weightlessness. Our a = g/5 gives an intermediate reduction to 48 kg.

Why Other Options Are Wrong:

(b) ignores acceleration. (c) would require upward acceleration. (d) and (e) correspond to different accelerations (g/10 and g/6 respectively), not the given value.

Common Pitfalls:

Confusing mass with weight units; forgetting that the scale reads the normal reaction; sign errors when the elevator moves downward.

Final Answer:

48 kg