Answer :
The acceleration of the elevator when the scale reads 400 N is 0.18m/[tex]s^{2}[/tex] downwards.
The acceleration of the elevator when the scale reads 610 N is 0.04m/[tex]s^{2}[/tex] upwards.
Yes, the student should worry because the elevator is in free fall, that is, there are no supporting cables on the elevator.
The tension in the cable for part A is 450 N.
The tension on the cable for part D is 0.
The given parameters;
- weight of the student, W = 570 N
- mass of the student and the elevator, m = 900 kg
When the scale of the elevator reads 400 N, is moving downwards since the apparent weight is less than the real weight
R = mg + ma
400 = 570 + ma
400 - 570 = ma
- 170 = ma
a = [tex]\frac{-170}{m}[/tex] = [tex]\frac{-170}{900}[/tex] = 0.18m/[tex]s^{2}[/tex] , downwards
so, The acceleration of the elevator when the scale reads 400 N is 0.18m/[tex]s^{2}[/tex] downwards.
The scale of the elevator will read more when it is moving upwards;
610 = mg + ma
610 = 570 + ma
610 - 570 = ma
ma =40
a = [tex]\frac{40}{m}[/tex] = [tex]\frac{40}{900}[/tex] = 0.04m/[tex]s^{2}[/tex] , upwards
So, the acceleration of the elevator when the scale reads 610 N is 0.04m/[tex]s^{2}[/tex] upwards.
When the scale reads, the upward acceleration is equal to the acceleration due to gravity and the student will be in free fall.
R = m(g - a)
g = a
R = m(0)
R = 0
Thus, the student should worry because the elevator is in free fall, that is, there are no supporting cables on the elevator.
The tension in the cable for part A is 400 N.
The tension in the cable for part D is 0.
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