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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

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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!

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