Motion of an Elevator

 

An elevator of mass M is accelerated upwards by applying a force F. A mass m initially situated at a height

of 1m above the floor of the elevator is falling freely. It will hit the floor of the elevator after a time equal to

  • √(2M/F+mg)
  • √(2M/F-mg)
  • √2M/F
  • √2M/(F+Mg)

Discussion:

Acceleration of elevator ae= F/M ( in upward direction)

Acceleration due to gravity is in downward direction so acceleration of mass as= -g

Acceleration of mass with respect to elevator

= as-ae

=(F/M)+g

=(F+Mg)/M

We know,

s=ut+(1/2)at2

From the given problem, we have s=1m

so,

1=(F+Mg)t2/M

t=√2M/(F+Mg)

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