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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?

Respuesta :

Answer:

Final velocity of the elevator will be 4.453 m/sec

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

[tex]F=mg+ma[/tex]

So according to question

[tex]1.46mg=mg+ma[/tex]

0.46 mg = ma

a = 0.46 g

a = 0.46×9.8 = 4.508 [tex]m/sec^2[/tex]

Initial velocity of elevator is 0 m/sec

From third equation of motion

[tex]v_f^2=v_i^2+2as[/tex]

[tex]v_f^2=0^2+2\times 4.508\times 2.2[/tex]

[tex]v_f=4.453m/sec[/tex]

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

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