Respuesta :
(a) [tex]-1.78 m/s^2[/tex] (downward)
The equation of the forces acting on the student is:
[tex]N-mg = ma[/tex] (1)
where
N is the normal reaction of the scale on the student
mg is the weight of the student
a is the acceleration of the student
The scale reads 450 N, so this is the normal reaction:
N = 450 N
Also, we know that the weight of the student is:
[tex]mg = 550 N[/tex]
So we can find its mass:
[tex]m=\frac{mg}{g}=\frac{550}{9.8}=56.1 kg[/tex]
So now we can solve eq.(1) to find the acceleration:
[tex]a=\frac{N-mg}{m}=\frac{450-550}{56.1}=-1.78 m/s^2[/tex]
where the negative sign means the acceleration is downward.
(b) [tex]2.14 m/s^2[/tex], upward
Again, the equation of the forces is
[tex]N-mg = ma[/tex]
where this time, the reading of the scale (and so, the normal reaction) is
N = 670 N
Solving for the acceleration, we find
[tex]a=\frac{N-mg}{m}=\frac{670-550}{56.1}=2.14 m/s^2[/tex]
and the positive sign means the acceleration here is upward.
(c) Yes
Let's assume the scale is reading zero. In terms of forces, this means that the normal reaction on the student is zero:
N = 0
So the equation of the forces simply becomes
[tex]mg = ma[/tex]
Therefore the acceleration is
[tex]a=g=-9.8 m/s^2[/tex]
which means that the elevator is accelerating downward at [tex]9.8 m/s^2[/tex]: this means that the elevator is in free fall, so yes, the student should worry.
(d) (a) 6817 N
Let's now consider the equation of the forces on the elevator:
[tex]T-mg = ma[/tex] (2)
where this time:
T is the tension in the cable
[tex]mg=(850 kg)(9.8 m/s^2)=8330 N[/tex] is the weight of the elevator+student system
[tex]a=-1.78 m/s^2[/tex] is the acceleration
Solving for T,
[tex]T=mg+ma=8330+(850)(-1.78)=6817 N[/tex]
(d) (b) 10149 N
Here we can use the same equation
[tex]T-mg = ma[/tex]
where the only difference is that the acceleration is
[tex]a=2.14 m/s^2[/tex]
Solving the equation for T, we find
[tex]T=mg+ma=8330+(850)(2.14)=10149 N[/tex]
(d) (c) 0
Again, same equation
[tex]T-mg = ma[/tex]
But this time, the acceleration is
[tex]a=-9.8 m/s^2[/tex]
So, we find:
[tex]T=mg+ma=8330+(850)(-9.8)=0[/tex]
So, the tension in the cable is zero, since the elevator is in free fall.
The acceleration of the elevator when the scale reads 450 N is 0.12 m/s² downwards.
The acceleration of the elevator when the scale reads 670 N is 0.14 m/s² upwards.
Yes, the student should worry because the elevator is under free fall, that is, there is no supporting cables on the elevator.
The tension of the cable for part a is 450 N
The tension on the cable for part c is 0.
The given parameters;
- weight of the student, W = 550 N
- mass of the student and the elevator, m = 850 kg
When the scale of the elevator reads 450 N, is moving downwards since the apparent weight is less than the real weight
R = mg + ma
450 = 550 + ma
450 - 550 = ma
- 100 = ma
[tex]a = \frac{-100}{m} = \frac{-100}{850} = 0.12 \ m/s^2[/tex] downwards
The scale of the elevator will read more when it is moving upwards;
670 = mg + ma
670 = 550 + ma
670 - 550 = ma
120 = ma
[tex]a = \frac{120}{850} = 0.14 \ m/s^2[/tex] upwards
When the scale reads, the upward acceleration is equal to the acceleration due to gravity and the student will be on free fall.
R = m(g - a)
g = a
R = m(0)
R = 0
Thus, the student should worry because elevator is under free fall, that is, there is no supporting cables on the elevator.
The tension of the cable for part a is 450 N
The tension on the cable for part c is 0.
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