Question

You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.59 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 3.7 m and then starts to slow down. What is the maximum speed of the elevator?

Answer 1

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

$v^2 = u^2+ 2 a s$         

$v =\sqrt{2a*s}$                    

  =$\sqrt(2\times 0.59\times 9.8\times 3.7)$

  = 6.54 m/s

maximum speed of the elevator is 6.54 m/s

Answer 2

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 = $9.8 m/s^{2}$

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:

$F_{net} = m(g + a)$

Thus

For maximum acceleration:

$1.5 mg = m(g + a)$

$1.5 g = g + a$

$a = 0.5 g = 0.5\times 9.8 = 4.9 m/s^{2}$

Now, from the third eqn of motion with initial velocity 0:

$v^{2} = 0 + 2ad$

$v = \sqrt{2\times 4.9\times 3.7} = 6.022 m/s$