Force of an elevator on person riding it

  • #1

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

Homework Equations



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
 
Last edited:

Answers and Replies

  • #2
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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
 
  • #3
PhanthomJay
Science Advisor
Homework Helper
Gold Member
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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.
 

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