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A physicist with a mass of 93 kg is standing on a scale in an elevator. While he is going up, the elevator is slowing down with an acceleration of [tex]4 \, \text{m/s}^2[/tex]. What does the scale show?

Answer :

Final answer:

The scale in an elevator slowing down at 4 m/s² with a physicist weighing 93 kg on it would show an apparent weight of approximately 539.4 N, which corresponds to about 55 kg.

Explanation:

The question involves a physicist with a mass of 93 kg standing on a scale in an elevator that is slowing down with an acceleration of 4 m/s² while moving upwards. To determine what the scale shows, we need to calculate the apparent weight of the physicist, which is the normal force exerted by the scale.

Firstly, we find the true weight of the physicist, which is the force due to gravity using the equation

F_gravity = m × g,

where m is the mass and g is the acceleration due to gravity (approximately 9.8 m/s²).

Then, since the elevator is slowing down (decelerating) while moving upward, the apparent weight felt by the physicist will be less than the true weight. This is calculated using the equation:

F_apparent = m × (g - a),

where a is the acceleration of the elevator.

Substituting the given values:

F_apparent = 93 kg × (9.8 m/s² - 4 m/s²)

F_apparent = 93 kg × 5.8 m/s²

F_apparent = 539.4 N

Therefore, the scale will show the equivalent mass that corresponds to an apparent weight of 539.4 N.

Since weight = mass × gravity, and we want to find the equivalent mass:

mass = weight / gravity.

The equivalent mass that the scale would show is:

mass = 539.4 N / 9.8 m/s²

mass ≈ 55 kg

So, the scale would show approximately 55 kg while the elevator is slowing down.

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