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Old Dec 12th 2010, 02:38 PM   #1
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Question AP Physics Elevator Problem

I'm really not sure if I'm doing this problem even remotely correct...

The Problem: An elevator starts from rest and after 3 seconds of constant acceleration reaches a velocity of 5 m/s going up. What is the normal force of the floor on a 110 kg person standing in the elevator.

My attempt: (I've bolded forces, subscripts lowercased)

v=u+at
5m/s=a(3s)
1.667 m/s^2 = a

Nep=Ffp-Fep
=ma+mg
=m(a+g)
=110kg(1.667-9.8)
=894.63N
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Old Dec 13th 2010, 01:16 AM   #2
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One question for you, do you feel heavier when going up an elevator or lighter?

Because in the solution you got, you mean to say that you feel lighter

Let's verify the forces on the person.

Normal Force - Weight = Net force

Where the normal force is upwards, the weight downwards and the net force upwards. g and a already have their correct signs.

F - mg = ma

F = m(g + a)

As the magnitude of acceleration is larger and you feel heavier when going up!
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