Elevator dynamics and apparent weight A 49 kg person stands on a spring scale in an elevator. Starting from rest, during the first 5 seconds the scale reading is 69 kg. Assuming constant acceleration, determine the elevator’s velocity at t = 5 s.

Introduction / Context:Apparent weight on a scale in an accelerating elevator reflects the normal reaction. From the reading we can back-calculate acceleration and then velocity. This is a standard inverse dynamics and kinematics application.

Given Data / Assumptions:

• Actual mass m = 49 kg.
• Scale shows apparent mass 69 kg during acceleration.
• Elevator starts from rest; duration t = 5 s; take g ≈ 9.8 m/s^2.
• Assume constant upward acceleration.

Concept / Approach:

Apparent weight N equals m(g + a) when accelerating upward. Scales calibrated in kilograms display N/g. Set N/g = 69 kg and solve for a. Then use v = u + a t with u = 0 to get velocity at 5 s.

Step-by-Step Solution:

Verification / Alternative check:

Using g = 9.81 yields a ≈ 4.00 m/s^2; velocity remains 20.0 m/s to three significant digits.

Why Other Options Are Wrong:

10 and 15 m/s underestimate, 25 and 30 m/s overestimate given the computed a = 4 m/s^2 for 5 s.

Common Pitfalls:

Confusing mass display (kg) with force; forgetting to divide by g; assuming downwards acceleration when the reading is higher than true weight.

Final Answer:

20 m/s