Physics 62 Online
OpenStudy (anonymous):

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

OpenStudy (anonymous):

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 $a = \frac{W}{ \frac{W_0}{g} } - g = g( \frac{W}{W_0} - 1)$

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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