Answer :
Answer:
The elevator's velocity after 6 seconds is 3.189m/s
Explanation:
Using Newton's 2nd law:Fnet = ma
Mg - N = ma
Where m = mass
N = normal force
a = acceleration
g = acceleration due to gravity
Substituting the given values and finding acceleration
(95kg)(9.8m/s^2) - 830N = 95kg x a
a = (931 -830) / 95
a = 101/95
a = 1.063m/s^2
After 3 seconds, the weight 0f the person is equal to the actual weight of the person, thus the elevator will be moving at a constant velocity.
Using kinematic equation:
V = u + at
Where V = final velocity
U = initial velocity = 0
a = acceleration
t = time
V = 0 + 1.063 × 3
V = 3.189m/s
This velocity does not change . The elevator travels the test of the time at the same velocity. Therefore,the velocity at 6 seconds = 3.189m/s
The motion of an elevator accelerating downwards, makes the passengers feel lighter, and when accelerating upwards, the passengers feel heavier
The elevator's velocity 6.0 s after starting is approximately 3.283 m/s
The reason the above value above value is correct is given as follows:
The known parameters:
Henry's mass, m = 95 kg
After the elevator starts moving;
The reading on the scale for the first 3.0 seconds = 830 N
The reading on the scale for the next 3.0 seconds = 930 N
Required:
The velocity of the elevator 6.0 s after starting
Solution:
Weight = mass × Acceleration due to gravity
Henry's weight, W = 95 kg × 9.8 m/s² ≈ 930 N
The apparent weight in downward motion is less than the weight
[tex]F_{scale} = 830 \, N[/tex]
With regards to the weight, W, we have;
[tex]F_{scale} = W - m\cdot a = m \cdot ( g- a)[/tex]
Therefore;
In the first 3 seconds, we have;
[tex]a_{(0\, s \ to \ 3 \, s)} = g - \dfrac{F_{scale}}{m} = 9.81 \ m/s^2 - \dfrac{830 \, N}{95 \, kg} \approx 1.073 \ m/s^2[/tex]
[tex]a_{(0\, s \ to \ 3 \, s)} \approx 1.073 \ m/s^2[/tex]
In the next 3 seconds, we have;
[tex]a_{(3\, s \ to \ 6 \, s)} = 9.81 \ m/s^2 - \dfrac{930 \, N}{95 \, kg} \approx 0.021\ m/s^2[/tex]
[tex]a_{(3\, s \ to \ 6 \, s)} \approx 0.021\ m/s^2[/tex]
Which gives;
v = u + a·t
u = 0
The velocity after the first 3 seconds, v₃, is given as follows;
v₃ = 0 + 1.073 m/s² × 3.0 s ≈ 3.22 m/s
v₃ ≈ 3.22 m/s
The velocity after the next 3 seconds, (t = 3), v₆, is given as follows;
v₆ = v₃ + [tex]a_{(3\, s \ to \ 6 \, s)}[/tex] × 3
∴ v₆ ≈ 3.22 + 0.021 × 3 = 3.283
The elevator's velocity 6.0 s after starting, v₆ ≈ 3.283 m/s
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