A 91 kg man stands on a spring scale in an elevator. Starting from rest, the eleva- tor ascends, attaining its maximum speed of 1.7 m/s in 1.07 s. It travels with this constant speed for the next 5 s, undergoes a uniform negative acceleration for 1.67 s, and comes to rest.
Scale
What does the spring scale register before the elevator starts to move?
The acceleration of gravity is 9.8 m/s2 . Answer in units of N.
009 (part 2 of 4) 10.0 points
What does the spring scale register during the first 1.07 s?
Answer in units of N.
010 (part 3 of 4) 10.0 points
What does the spring scale register while the elevator is traveling at constant speed?
Answer in units of N.
011 (part 4 of 4) 10.0 points
What does the spring scale register during the negative acceleration?
Before the elevator starts to move, the spring scale registers the weight of the man, which we can calculate using the equation W = mg, where m is the mass of the man and g is the acceleration due to gravity.
So, W = 91 kg * 9.8 m/s^2 = 892.8 N
During the first 1.07 s, the elevator is accelerating and the spring scale will register a greater force than just the man's weight. However, since no actual value is given for this acceleration, we cannot calculate the exact force registered by the scale.
While the elevator is traveling at a constant speed, there is no acceleration, and therefore, the spring scale will only register the weight of the man. The force registered by the scale will be the same as calculated before, which is 892.8 N.
During the negative acceleration, the spring scale will register a reduced force as the elevator decelerates. The exact force registered will depend on the magnitude of the negative acceleration, which is not given in the question. So, we cannot determine the exact force registered by the scale during this period.
To determine what the spring scale registers at different points in time, we can use the concept of force and Newton's second law of motion.
1. Before the elevator starts to move:
The spring scale registers the man's weight, which is the force of gravity acting on him. The weight can be calculated using the formula W = m * g, where W is the weight, m is the mass, and g is the acceleration due to gravity. In this case, the mass is 91 kg and the acceleration due to gravity is 9.8 m/s^2. Therefore, the weight is:
W = (91 kg) * (9.8 m/s^2) = 892.8 N.
2. During the first 1.07 s:
The spring scale registers the force exerted on the man due to the upward acceleration of the elevator. To calculate this force, we can use Newton's second law: F = m * a, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration is given as 1.7 m/s^2 and the mass is 91 kg. Therefore, the force is:
F = (91 kg) * (1.7 m/s^2) = 154.7 N.
3. While the elevator is traveling at a constant speed:
The spring scale registers the man's weight, which remains constant as there is no acceleration. The weight is the same as calculated in the first step:
W = 892.8 N.
4. During the negative acceleration:
The spring scale registers the difference between the man's weight and the force required to decelerate him. Since the acceleration is negative, we can use the formula:
F = m * a, where F is the force, m is the mass, and a is the acceleration. In this case, the acceleration is given as -1.7 m/s^2 and the mass is 91 kg. Therefore, the force is:
F = (91 kg) * (-1.7 m/s^2) = -154.7 N.
To answer these questions, we need to consider the forces acting on the man in different situations.
1. Before the elevator starts to move:
When the elevator is at rest, the only force acting on the man is the force of gravity. The scale registers the force of gravity, which can be calculated using the formula F = mg, where m is the mass of the man and g is the acceleration due to gravity. Plugging in the given values, we can calculate the force of gravity.
F = (91 kg)(9.8 m/s^2) = 892.8 N
Therefore, the spring scale will register a force of 892.8 Newtons before the elevator starts to move.
2. During the first 1.07 s:
During this period, the elevator is accelerating upwards. The net force acting on the man can be calculated using Newton's second law, F = ma, where F is the net force, m is the mass of the man, and a is the acceleration. In this case, the acceleration can be found using the formula a = Δv / Δt, where Δv is the change in velocity and Δt is the time taken. Plugging in the given values, we can calculate the acceleration.
a = (1.7 m/s) / (1.07 s) ≈ 1.5888 m/s^2
Now, we can calculate the net force.
F = (91 kg)(1.5888 m/s^2) ≈ 144.686 N
Therefore, the spring scale will register a force of approximately 144.686 Newtons during the first 1.07 seconds.
3. While the elevator is traveling at constant speed:
When the elevator is traveling at a constant speed, it means there is no net force acting on the man. This means that the force of gravity is balanced by the normal force from the scale, resulting in a zero net force. Therefore, the spring scale will register a force of zero Newtons while the elevator is traveling at a constant speed.
4. During the negative acceleration:
During this period, the elevator is decelerating or undergoing negative acceleration. The net force acting on the man can be calculated similarly to the previous situations. The acceleration can be found using the formula a = Δv / Δt again, but this time the change in velocity will be negative since the elevator is slowing down. Plugging in the given values, we can calculate the acceleration.
a = (-1.7 m/s) / (1.67 s) ≈ -1.01796 m/s^2
Now, we can calculate the net force.
F = (91 kg)(-1.01796 m/s^2) ≈ -92.66376 N
Therefore, the spring scale will register a force of approximately -92.66376 Newtons during the negative acceleration.
To summarize:
- Before the elevator starts to move, the spring scale will register a force of 892.8 N.
- During the first 1.07 s, the spring scale will register a force of approximately 144.686 N.
- While the elevator is traveling at constant speed, the spring scale will register a force of 0 N.
- During the negative acceleration, the spring scale will register a force of approximately -92.66376 N.
if a = 0
F = m g = (91)(9.8) = 892 N
a = 1.7 m/s / 1.07 s = 1.59 m/s^2
Fup - mg down = m a = 91(1.59)
Fup = 892+145 = 1037 N
constant speed, no acceleration so back to
m g = 892
892 -91(1.67/t)
and
0 = 1.7 - 1.67 t
t = 1.02 seconds to stop
so
892 - 91 (1.67/1.02)
= 743 N