What will a spring scale read for the weight of a 50 kg woman in an elevator that moves as follows?
(a) upward with constant speed of 3.0 m/s
Enter a number.1 N
(b) downward with constant speed of 3.0 m/s
Enter a number.2 N
(c) upward with acceleration of 0.35g
Enter a number.3 N
(d) downward with acceleration 0.35g
Enter a number.4 N
(e) in free fall
(e) In free fall, the spring scale will read zero since there is no support force acting on the woman's body due to gravity.
To determine the readings of a spring scale in different scenarios, we need to consider the forces acting on the woman. The weight of an object is the force exerted on it due to gravity and can be calculated using the formula:
Weight = mass × acceleration due to gravity
Here, the mass of the woman is 50 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
(a) When the elevator moves upward with a constant speed of 3.0 m/s, the woman experiences no acceleration. Therefore, the net force acting on her is zero, and the spring scale will read the same weight as when she is at rest on the ground.
The weight of the woman is given by:
Weight = 50 kg × 9.8 m/s^2 ≈ 490 N
So, the spring scale will read approximately 490 N.
(b) When the elevator moves downward with a constant speed of 3.0 m/s, the situation is the same as in scenario (a), and the spring scale will read the weight of the woman, which is approximately 490 N.
(c) When the elevator moves upward with an acceleration of 0.35g, we need to consider the additional force acting on the woman due to this acceleration. The net force acting on the woman is the sum of the force due to gravity and the force due to the acceleration of the elevator. The weight of the woman remains the same:
Weight = 50 kg × 9.8 m/s^2 ≈ 490 N
The additional force due to the acceleration is given by:
Force = mass × acceleration = 50 kg × 0.35g ≈ 171.5 N
Therefore, the total force acting on the woman is the sum of the force due to gravity and the force due to acceleration:
Total force = Weight + Force = 490 N + 171.5 N ≈ 661.5 N
So, the spring scale will read approximately 661.5 N.
(d) In the scenario where the elevator moves downward with an acceleration of 0.35g, we follow the same procedure as in scenario (c). The weight of the woman remains the same:
Weight = 50 kg × 9.8 m/s^2 ≈ 490 N
The additional force due to the acceleration is given by:
Force = mass × acceleration = 50 kg × 0.35g ≈ 171.5 N
The total force acting on the woman is the difference between the force due to gravity and the force due to acceleration:
Total force = Weight - Force = 490 N - 171.5 N ≈ 318.5 N
So, the spring scale will read approximately 318.5 N.
(e) In free fall, the woman experiences zero gravity, which means she is weightless. In this scenario, the spring scale will read zero because there is no force acting on the woman.
Therefore, the spring scale readings for each scenario are as follows:
(a) 490 N
(b) 490 N
(c) 661.5 N
(d) 318.5 N
(e) 0 N
this is easy and immposible at the same time. i'm not a smart indiviual so i wouldent know much. all i know is that i have the physics final tomorrow and this and im gonna fail..
the weight of the woman is mg
When going up, the scale will read mg+ma
when going down, the scale will read mg-ma