A Monash College student of mass70 kg standing in an elevator briefly experiences a constant downward vertical speed of 2 m/s as he moves from level 6 to level 2. What is apparent weight of the student at this moment?

Question

BlackTom AI · Accepted Answer

We start by identifying the physical situation: a student in an elevator moving at a constant downward vertical speed. The key concept is Newton's second law for non-accelerating motion along the elevator frame: if there is no acceleration, the support (normal) force equals the true weight.

Option a: 'Greater than the gravitational weight force' would imply the elevator (or the contact force from the floor) is exerting more force on the student than gravity does. That would require a net upward acceleration of the student relative to the elevator frame, which contradicts the given condition of constant velocity (zero acceleration). Therefore this option is incorrect.

Option b: 'Less than the gravitational weight force' would imply a net downward unbalanced force causing downward acceleration, which again contradicts the stated constant speed (no acceleration). Since the acceleration is zero, the floor cannot be supporting less than the weight. Hence this option is incorrect.

Option c: 'Equal to the gravitational weight force' means the normal force from the floor exactly balances gravity, producing zero acceleration in the elevator frame. This aligns with the condition of constant velocity (no acceleration) and is the expected apparent weight when there is no acceleration. Therefore this option is correct.

Additional note: The actual numerical value would be m g, which for a 70 kg mass is about 70 × 9.8 ≈ 686 N, but the question focuses on the comparison to gravitational weight rather than the numeric value.

A Monash College student of mass70 kg standing in an elevator briefly experiences a constant downward vertical speed of 2 m/s as he moves from level 6 to level 2. What is apparent weight of the student at this moment?

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