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

Question

BlackTom AI · Accepted Answer

In this scenario, the elevator is moving at a constant upward velocity, which means its acceleration is zero.

Option a: Less than the gravitational weight force. This would occur if the elevator were accelerating downward or if there were a downward net force; neither happens here since acceleration is zero, so this statement is incorrect.

Option b: Greater than the gravitational weight force. A greater apparent weight would require a positive upward acceleration of the elevator, creating an additional upward force on the student. Again, the acceleration is zero, so this cannot be true.

Option c: Equal to the gravitational weight force. With zero acceleration, the normal force (which is the apparent weight felt as the contact force from the floor) exactly balances the gravitational force mg pulling downward. Therefore, the apparent weight equals mg. Calculating mg: m = 70 kg, g ≈ 9.8 m/s^2, so mg ≈ 70 × 9.8 = 686 N. Hence, the apparent weight is equal to the true gravitational weight, about 686 N.

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

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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?

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

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