Respuesta :
Answer:
maximum speed of the elevator is 6.54 m/s
Explanation:
given,
force exerted on the elevator is 1.59 times passenger weight.
when the elevator accelerates upward with rate of a.
total force
F = m × (a+g)
1.59× mg = ma + mg
0.59 mg = ma
a = 0.59 g
we also have that
[tex]v^2 = u^2+ 2 a s[/tex]
[tex]v =\sqrt{2a*s}[/tex]
=[tex]\sqrt(2\times 0.59\times 9.8\times 3.7)[/tex]
= 6.54 m/s
maximum speed of the elevator is 6.54 m/s
Answer:
The maximum speed is 6.022 m/s
Solution:
As per the question:
Distance covered by the elevator in the upward direction, d = 3.7 m
Maximum force exerted by the elevator on the passenger = 1.59 w N
where
w = weight of the passenger = mg
where
m = mass of the passenger
g = acceleration due to gravity = [tex]9.8 m/s^{2}[/tex]
Therefore,
Maximum force exerted by the elevator on the passenger = 1.59mg N
Now, for the maximum speed of the elevator:
Net force on the floor of the elevator when it moves upwards:
[tex]F_{net} = m(g + a)[/tex]
Thus
For maximum acceleration:
[tex]1.5 mg = m(g + a)[/tex]
[tex]1.5 g = g + a[/tex]
[tex]a = 0.5 g = 0.5\times 9.8 = 4.9 m/s^{2}[/tex]
Now, from the third eqn of motion with initial velocity 0:
[tex]v^{2} = 0 + 2ad[/tex]
[tex]v = \sqrt{2\times 4.9\times 3.7} = 6.022 m/s[/tex]
