Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 3.0 s after the elevator starts moving, then 930 N for the next 3.0 s. What is the elevator’s velocity 6.0 s after starting?

The motion of an elevator accelerating downwards, makes the passengers feel lighter, and when accelerating upwards, the passengers feel heavier

The elevator's velocity 6.0 s after starting is approximately 3.283 m/s

The reason the above value above value is correct is given as follows:

The known parameters:

Henry's mass, m = 95 kg

After the elevator starts moving;

The reading on the scale for the first 3.0 seconds = 830 N

The reading on the scale for the next 3.0 seconds = 930 N

Required:

The velocity of the elevator 6.0 s after starting

Solution:

Weight = mass × Acceleration due to gravity

Henry's weight, W = 95 kg × 9.8 m/s² ≈ 930 N

The apparent weight in downward motion is less than the weight

With regards to the weight, W, we have;

Therefore;

In the first 3 seconds, we have;

In the next 3 seconds, we have;

Which gives;

v = u + a·t

u = 0

The velocity after the first 3 seconds, v₃, is given as follows;

v₃ = 0 + 1.073 m/s² × 3.0 s ≈ 3.22 m/s

v₃ ≈ 3.22 m/s

The velocity after the next 3 seconds, (t = 3), v₆, is given as follows;

v₆ = v₃ + × 3

∴ v₆ ≈ 3.22 + 0.021 × 3 = 3.283

The elevator's velocity 6.0 s after starting, v₆ ≈ 3.283 m/s

Learn more about motion in an elevator here:

link

true tan speed = ang vel x radius

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