A 65-kg woman descends in an
elevator that briefly accelerates at
0.2g downward. She stands on a
scale that reads in kg.
• During this acceleration, what is her
weight and what does the scale
read?
• What does the scale read when the
elevator descends at a constant
speed of 2m/s?
During the acceleration, the woman's weight is 65 kg and the scale reads 65 kg. When the elevator descends at a constant speed of 2m/s, the scale reads 65 kg.
To determine the weight and scale reading of the woman during the elevator's acceleration, we need to consider the force acting on her.
1. Weight and scale reading during acceleration:
The weight of an object is given by the formula W = mg, where m is the mass of the object and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given:
Mass of the woman (m) = 65 kg
Acceleration due to gravity (g) = 9.8 m/s^2
The elevator briefly accelerates downward at 0.2g, so the acceleration experienced by the woman is 0.2 * 9.8 m/s^2 = 1.96 m/s^2 (downward).
a) Weight during acceleration:
W = mg = 65 kg * 9.8 m/s^2 = 637 N (upward)
During acceleration, the apparent weight of the woman is equal to the sum of her actual weight and the weight due to the acceleration force.
b) Scale reading during acceleration:
The scale measures the normal force exerted by the woman on it. Since the elevator is accelerating downward, the normal force will be less than the woman's actual weight. Therefore, the scale reading will be less than her weight during this acceleration.
2. Scale reading during constant speed descent:
When the elevator descends at a constant speed, it implies that there is no net acceleration acting on the woman. Therefore, the scale reading will be equal to her weight.
Using the same formula as before:
W = mg = 65 kg * 9.8 m/s^2 = 637 N (upward)
So, the scale reading when the elevator descends at a constant speed of 2 m/s will be 637 N, which is equal to her weight.
To summarize:
a) During acceleration:
- Weight: 637 N (upward)
- Scale reading: less than 637 N
b) During constant speed descent:
- Weight: 637 N (upward)
- Scale reading: 637 N (upward)
To determine the answers to these questions, we need to consider the forces acting on the woman in each scenario. Let's break it down step by step.
1. During acceleration at 0.2g downward:
First, we need to calculate the woman's weight. Weight is the force of gravity acting on an object and is determined by multiplying the mass of the object by the acceleration due to gravity (9.8 m/s^2).
Weight = mass * acceleration due to gravity
Weight = 65 kg * 9.8 m/s^2
Weight = 637 N
However, during this acceleration, there is an additional force acting on the woman, which is the acceleration of the elevator downward. The acceleration of the elevator is given as 0.2g, where g = acceleration due to gravity.
The net force acting on the woman during this acceleration is the sum of her weight and the downward force exerted by the elevator.
Net force = weight + force exerted by the elevator
Net force = 637 N - (0.2 * 65 kg * 9.8 m/s^2)
The scale measures the force exerted on it, so it will read the net force acting on the woman. Hence, the scale reading would be equal to the net force calculated above.
2. During a constant speed descent of 2 m/s:
When the elevator descends at a constant speed, the woman is not accelerating. Therefore, the net force acting on her is zero.
Since the net force is zero, the scale will only measure her weight without any additional force due to acceleration. Thus, the scale reading would be equal to her weight, which is 637 N.
In summary:
1. During acceleration at 0.2g downward, the woman's weight is 637 N, and the scale will read the net force acting on her, which is calculated as explained above.
2. During a constant speed descent of 2 m/s, the woman's weight is still 637 N, and the scale will read her weight without any additional force due to acceleration.