A 75 kg man stands on a spring scale in an elevator. Starting from rest, the elevator ascends, attaining its maximum speed of 1.6 m/s in 1.1 s. It travels with this constant speed for the next 7.2 s, undergoes a uniform
negative acceleration for 2.4 s, and comes to rest.
What does the spring scale register before the elevator starts to move?
The acceleration of gravity is 9.8 m/s^2
75*9.8
since nothing is moving yet, only gravity is involved.
To determine what the spring scale registers before the elevator starts to move, we need to understand the concept of apparent weight in an elevator.
When an elevator is at rest or moving at a constant velocity, the apparent weight of an object inside it is equal to its actual weight. However, when the elevator is accelerating or decelerating, the apparent weight of the object will differ from its actual weight due to the effects of the acceleration.
In this case, since the elevator is starting from rest and ascending, we can assume that it undergoes an upward acceleration. To find the magnitude of this acceleration, we can use the equation:
Final velocity (v) = Initial velocity (u) + Acceleration (a) * Time (t)
Given:
Initial velocity (u) = 0 m/s
Final velocity (v) = 1.6 m/s
Time (t) = 1.1 s
Using the formula, we can rearrange it to solve for acceleration (a):
a = (v - u) / t
Substituting the given values:
a = (1.6 - 0) / 1.1
a = 1.6 / 1.1
a ≈ 1.45 m/s^2
Next, we need to calculate the apparent weight of the man before the elevator starts to move. The apparent weight is equal to the actual weight plus the force exerted by the acceleration.
The equation to find the apparent weight is:
Apparent weight = Actual weight + (Mass * Acceleration)
Given:
Mass of the man = 75 kg
Acceleration due to gravity (g) = 9.8 m/s^2
Apparent weight = 75 kg * (9.8 m/s^2 + 1.45 m/s^2)
Apparent weight = 75 kg * 11.25 m/s^2
Apparent weight ≈ 843.75 N
Therefore, the spring scale will register an apparent weight of approximately 843.75 N before the elevator starts to move.