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 1.46 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 2.2 m and then starts to slow down.What is the maximum speed of the elevator?

Answer 1

Final velocity of the elevator will be 4.453 m/sec

Step-by-step explanation:

Let mass is m

Acceleration due to gravity is g m/sec^2

Distance s = 2.2 m

As the elevator is moving upward so net force on elevator

`F=mg+ma`

So according to question

`1.46mg=mg+ma`

0.46 mg = ma

a = 0.46 g

a = 0.46×9.8 = 4.508
`m/sec^2`

Initial velocity of elevator is 0 m/sec

From third equation of motion

`v_f^2=v_i^2+2as`

`v_f^2=0^2+2* 4.508* 2.2`

`v_f=4.453m/sec`

So final velocity of the elevator will be 4.453 m/sec