Question

Mr. Bates steps into an elevator at the top floor of his hotel. He takes the elevator down to get out of the building. If Mr. Bates weighs 834 N and the elevator is accelerating down at 2 m/s, what is his apparent weight?

Answer 1

Mr. Bates's apparent weight in an accelerating elevator is found by subtracting the product of his mass and the elevator's downward acceleration from his actual weight. Due to the acceleration, his apparent weight is less than his gravitational weight of 834 N.

Step-by-step explanation:

The student asked what Mr. Bates's apparent weight would be in an elevator accelerating downwards at 2 m/s² if his actual weight is 834 N. To solve this, we need to use the concept of apparent weight which is the normal force provided by the scale in an accelerating system.

To find the apparent weight, we use the equation Fs = ma + mg, where Fs is the scale force (apparent weight), m is the mass of the person, a is the acceleration of the elevator, and g is the acceleration due to gravity (9.8 m/s²). In this scenario, the elevator is accelerating down, so the acceleration a will be negative relative to g.

Since the actual weight W is the gravitational force, we can express it as W = mg. Therefore, the mass m of Mr. Bates is m = W/g, which we can calculate by dividing his weight by the acceleration due to gravity. To find Mr. Bates's apparent weight, we subtract the product of his mass and the negative acceleration from his actual weight (W - ma).

The scale force will therefore be less than the actual weight because the elevator is accelerating downwards, reducing the normal force. If the acceleration were upwards, the apparent weight would be greater due to additional force required to accelerate Mr. Bates upwards.