Question

The upward normal force exerted by the floor is 620N on an elevator passenger who weighs 650 N.

How do i find the acceleration

The upward normal force exerted by the floor is 620N on an elevator passenger who weighs 650 N.

How do i find the acceleration
@Physics

Answer 1

The weight of a passenger in an accelerating elevator is
\[W = m(g+a)\]
so
\[a = \frac{W}{m} - g\]
in your problem, you should note that \[m = W_0/g\] where \[W_0\] is his normal weight. That means that
\[a = \frac{W}{ \frac{W_0}{g} } - g = g( \frac{W}{W_0} - 1) \]

Answer 2

I should have mentioned that
\[W\] denotes the apparent weight in the elevator, in your case 620 Newtons.

Answer 3

cool using that i got .4523 m/s^2
thanks!