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?

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

  =

  = 6.54 m/s

maximum speed of the elevator is 6.54 m/s

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 =

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:

Thus

For maximum acceleration:

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

mistakes are ok