# Force of an elevator on person riding it

• captainjack
In summary, an elevator with a mass of 48.0 kg accelerates upward from rest at the ground floor, 5.00 m below the first floor, and passes the second floor 1.65 s later. The normal force exerted by the floor of the elevator on a 48.0 kg passenger can be calculated by using the equations N=m(g+a) and v^2 = 2*a*s, with the given values of mass, acceleration, and distance between floors. The equations also require the use of another equation that utilizes the given time. After solving for the acceleration, the normal force can be calculated as N=48(9.8+0.630)=514.46 N.

## 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 don't 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

Last edited:
captainjack said:
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 don't 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

There are 3 speeds of interest, V_0 is 0 at the ground floor. V1 is the speed at the first floor. V2 is the speed at the 2nd floor. See if you can now make sense of the equations. But you missed a step somewhere..you have to use another equation , that makes use of the given time, before you can solve for a.