Solving for weight and mass in kg in an elevator

  • #1
42
1
Homework Statement:
A 65 kg woman descends in an elevator that briefly accelerates at 0.20 g downward. She stands on a scale that reads in kg. During this acceleration, what is her weight and what does the scale read?
Relevant Equations:
Use Weight= mass*gravitational acceleration and F= ma
Fn= normal force

W=mg
W=(65)(9.8)= 637N

Fn-637= -(65)(0.2*9.8)
Fn= 509.6N

I got the weight of the woman in the elevator, but I have trouble finding what the scale reads in kg. I was given the solution which read:
a = (g-0.2g)= 0.8g
W= 0.8g*509.6N= 52kg.

My questions
1. why do I have to subtract 0.2g from 1g?
2. Why don't they substitute gravitational constant(9.8m/s^2) in "g"?
 

Answers and Replies

  • #2
Delta2
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Homework Statement: A 65 kg woman descends in an elevator that briefly accelerates at 0.20kg downward. She stands on a scale that reads in kg. During this acceleration, what is her weight and what does the scale read?
Homework Equations: Use Weight= mass*gravitational acceleration and F= ma

Fn= normal force

W=mg
W=(65)(9.8)= 637N

Fn-637= -(65)(0.2*9.8)
Fn= 509.6N

I got the weight of the woman in the elevator, but I have trouble finding what the scale reads in kg.
You are correct so far. To convert the force to mass you just divide by g and thus you find what the scale reads in Kg.
I was given the solution which read:
a = (g-0.2g)= 0.8g
W= 0.8g*509.6N= 52kg.

My questions
1. why do I have to subtract 0.2g from 1g?
2. Why don't they substitute gravitational constant(9.8m/s^2) in "g"?
1. IF we follow exactly the same route as your solution but replace the numbers in the end we' ll have
$$F_n-mg=-m(0.2g)\Rightarrow F_n=mg-m(0.2g)=m(g-0.2g)$$
So it's a matter of algebra and replace the numbers in the end for why we have to just subtract 0.2g from g and multiply that by the mass m to find the normal force.
2.Because we have to divide the force by g to convert the force ##F_n## to the Kg scale .
 
  • #3
42
1
I got it! Thanks a lot!
 

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