# Force of an elevator on person riding it

captainjack

## Homework Statement

An elevator accelerates constantly upward from rest and passes the second floor, 5.00 m above the first floor, 1.65 s after it passes the first floor. What force does the floor of the elevator exert on a 48.0 kg passenger? Assume the elevator starts from rest at the ground floor, 5.00m below the first floor

N force=mg+ma
v^2 = 2*a*s

## The Attempt at a Solution

I know that based on the free body diagram the normal force is mg +ma (so I just need to solve this to get the force of the floor on the person). I'm having trouble solving for the acceleration though. I've tried using the equation above (which I assumed was the correct one to use), but according to the sample midterm solutions that my professor gave me this isn't even the equation to use. According to the answer key...
V2^2-V1^2=2a*5
V1^2=2a*5
V2^2=20a=4.47√a
so a=0.630m/s^2 (I kept getting 0.918)
but what I dont understand is how you would get all of these numbers. Isn't V1 supposed to be 0 because its starting from rest? ...which would make N=m(g+a)=48(9.8+0.919)=514.46

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## Answers and Replies

cupid.callin
V2^2-V1^2=2a*5
V1^2=2a*5
V2^2=20a=4.47√a
so a=0.630m/s^2 (I kept getting 0.918)
but what I dont understand is how you would get all of these numbers. Isn't V1 supposed to be 0 because its starting from rest? ...which would make N=m(g+a)=48(9.8+0.919)=514.46

v1 is not 0 as it is velocity at first floor

v is 0 at ground floor