An elevator is moving down with an acceleration of #3.36# #m##/##s^2#. What would be the apparent weight of a #64.2# #kg# man in the elevator?

1 Answer

414.448 N

Explanation:

Apparent Weight = #m(g - a)# if the elevator is coming down.

Just use the formula to find it.

So Apparent Weight = #64.2 (9.8- 3.36)# N

#color(white)(xxxxxxxxxxxxx)#= #64.2 * 6.44# N

#color(white)(xxxxxxxxxxxxx)#= #413.448# N

Now I will Explain what is Happening Inside the elevator.

http://2.bp.blogspot.com/-OnVwTw0q-OY/Uwxnls6gb8I/AAAAAAAAENM/51UG9BYjhcI/s1600/lift-07png

While moving down,

The weight of the man = mg
And the acceleration of the lift = a;

So, Force exerted on the lift = ma; [Here m is the total mass of the man and the elevator]

It is not falling freely, so the a must be less than g.

So, Some Force is restricting the free fall. The force is Normal Reaction.

So, Normal Reaction = R = mg - ma = m(g-a)

Which is the weight man will feel inside the elevator.

Hence explained my answer.