Respuesta :
Answer:
82.39 kg
72 kg
62.07 kg
Explanation:
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{1.7-0}{1.2}\\\Rightarrow a=1.4167\ m/s^2[/tex]
Force on the scale would be
[tex]F=m(g+a)\\\Rightarrow F=72(9.81+1.4167)\\\Rightarrow F=808.3224\ N[/tex]
[tex]808.3224=m\times 9.81\\\Rightarrow m=\frac{808.3224}{9.81}\\\Rightarrow m=82.39\ kg[/tex]
The scale would register 82.39 kg when going up
When the speed becomes constant the acceleration becomes 0. So, the scale would show 72 kg in the five seconds
[tex]v=u+at\\\Rightarrow a=\frac{v-u}{t}\\\Rightarrow a=\frac{0-1.7}{1.8}\\\Rightarrow a=-0.944\ m/s^2[/tex]
Force on scale
[tex]F=m(g+a)\\\Rightarrow F=72(9.81-0.944)\\\Rightarrow F=638.352\ N[/tex]
[tex]638.352=m\times 9.81\\\Rightarrow m=\frac{638.352}{9.81}\\\Rightarrow m=65.07\ kg[/tex]
The scale would show 65.07 kg while going down.
To solve the problem we must know about the concept of Accleration.
What is acceleration?
Acceleration is defined as the rate of change of velocity with respect to time. It is given by the formula,
[tex]a=\dfrac{v-u}{t}[/tex]
What does the spring scale register before the elevator starts to move?
As the elevator is not moving up or down, therefore there will be an acceleration in the spring scale, therefore, the spring scale will measure the weight of the person.
The measure in the spring scale = the weight of the person
= 72 x 9.81
= 70.632 N
What does it register during the first 1.20 s?
To find the measure in the first 1.20 s, when the elevator starts to move we need to calculate the acceleration through which the elevator must be going,
Given to us
Initial velocity of the elevator, u = 0 m/s
Final velocity of the elevator, v = 1.70 m/s
Time taken by the elevator to reach the maximum velocity, t = 1.20 s
Mass of the person, m = 72 kg
Substitute the value in the formula of acceleration,
[tex]a=\dfrac{v-u}{t}\\\\a=\dfrac{1.70-0}{1.20}\\\\a = 1.4167\rm\ m/s^2[/tex]
The weight that will be measured by the spring scale,
[tex]F = m(g+a)\\\\F = 72(9.81+1.4167)\\\\F = 808.32\rm\ N[/tex]
Thus, the measure that will be done by the spring scale in the first 1.20 s is 808.32 N.
What does it register while the elevator is traveling at a constant speed?
As the elevator is moving with constant speed, therefore there will be no acceleration in the spring scale, therefore, the spring scale will measure the weight of the person.
The measure in the spring scale = the weight of the person
= m x g
= 72 x 9.81
= 70.632 N
Learn more about Acceleration:
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