A Monash College student of mass 77.8 kg standing in an elevator briefly experiences a constant upward acceleration of 1.5 as he moves from level 2 to level 6. What is apparent weight of the student at this moment?

Question

BlackTom AI · Accepted Answer

Apparent weight equals the normal force from the floor (N). From Newton’s second law in the vertical direction: N − mg = ma, so N = m(g + a).

Given m = 77.8 kg, a = 1.5 m/s^2 and using g = 9.8 m/s^2 (standard value used here):
1. Compute effective acceleration: g + a = 9.8 + 1.5 = 11.3 m/s^2.
2. Multiply by the mass: N = 77.8 × 11.3.

Break the multiplication into parts:
- 77.8 × 11 = 77.8 × (10 + 1) = 778.0 + 77.8 = 855.8
- 77.8 × 0.3 = 23.34
- Sum: 855.8 + 23.34 = 879.14 N

So the apparent weight (normal force) is 879.14 N.

Why this is correct: when the elevator accelerates upward, the floor must push up harder than the static weight mg to produce the upward acceleration; mathematically that extra push is m·a, so N = mg + m a.

Why the other choices are incorrect:
- 879.92 N would result from using g = 9.81 m/s^2 (77.8 × 11.31 ≈ 879.92) instead of g = 9.8.
- 646.52 N corresponds to N = m(g − a) with g = 9.81 (downward or decelerating upward motion), not upward acceleration.
- 116.70 N is just m·a (the net upward force), not the total normal force (apparent weight).
- 763.22 N is the normal force when stationary (mg with g = 9.81), not when accelerating upward.

Confirming final answer: 879.14 N (matches the provided correct answer).

A Monash College student of mass 77.8 kg standing in an elevator briefly experiences a constant upward acceleration of 1.5 as he moves from level 2 to level 6. What is apparent weight of the student at this moment?Numerical

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