calculusxy
  • calculusxy
MEDAL!!! How much work is done by an elevator that carries a person with a mass of 70 kg up 40 meters? Assume the elevator has a constant acceleration of 2 m/s^2.
Mathematics
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SOLVED
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chestercat
  • chestercat
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calculusxy
  • calculusxy
@Michele_Laino
Michele_Laino
  • Michele_Laino
The elevator is not a inertial system, nevertheless, we can consider it, as an inertial system if we introduce the fictitious force -m*Ac where Ac is the ecceleration of the elevator, and m is the mass of the person inside the elevator. So the situation is like below: |dw:1437833008631:dw|
Michele_Laino
  • Michele_Laino
in other word, on the person are acting two forces: the weight force whose magnitude is m*g= 70*9.81=...newtons the fictitious force, whose magnitude is m*Ac=70*2=...newtons Since the elevator is a inertial system, the laws of physics apply unchanged in it, so the requested work is: \[\Large W = m\left( {g + {A_c}} \right)h = ...joules\] where h=40 meters

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calculusxy
  • calculusxy
So we are learning about energy now and the formulas that we have gone through so far are: Power = w/t Work = f x d Potential Energy: mass x gravity x distance Kinetic Energy: 1/2mv^2
calculusxy
  • calculusxy
The teacher said that he incorporated these formulas into the questions, so i need to understand what formulas go to what question to get the answer.
Michele_Laino
  • Michele_Laino
you can apply this formula: Work = f x d providing that the force f is: f= m(Ac+g) where Ac=2 m/sec^2
calculusxy
  • calculusxy
Why can't we use the formulas to find the power because that this the amount of work done in an interval?
calculusxy
  • calculusxy
*formula
Michele_Laino
  • Michele_Laino
the elevator is not an inertial system, since there are 2 forces acting on the person inside the elevator. I think it is better if we use the formula: work = force * distance
calculusxy
  • calculusxy
I still don't understand how you find the force.
Michele_Laino
  • Michele_Laino
In other words, I have applied the Principle of Equivalence, and I consider the elevator like an inertial system, providing that to introduce the fictitious force
calculusxy
  • calculusxy
i am sorry , but i don't know the principle of equivalence because i am just a rising eighth grader.
Michele_Laino
  • Michele_Laino
when you use an elevator in order to go up, at the starting of motion, you feel like pressed towards the floor of the elevator, right?
calculusxy
  • calculusxy
yes
Michele_Laino
  • Michele_Laino
that is due to the fictitious force which is acting on you
calculusxy
  • calculusxy
ok
Michele_Laino
  • Michele_Laino
that fictitious force will add to your weight force
calculusxy
  • calculusxy
ok
calculusxy
  • calculusxy
can we use the formula f = m x a?
Michele_Laino
  • Michele_Laino
so the total force which is acting on you is: fictitious force+weight force
Michele_Laino
  • Michele_Laino
and the work done by the elevator is the work done by that total force
calculusxy
  • calculusxy
\[Force = mass \times acceleration\] \[Force = 70 kg \times 2m/s^2\] \[Force = 140 N \] Work: \[Force \times Distance \] \[140N \times 40 m\] \[1600J\]
Michele_Laino
  • Michele_Laino
yes! the fictitious force is given by the subsequent formula: f=m*Ac where Ac=2 m/sec^2
Michele_Laino
  • Michele_Laino
now you have to add the work done by the weight force
calculusxy
  • calculusxy
Oh... i am sorry about misinterpreting it because i didn't know that Ac actually meant acceleration.
Michele_Laino
  • Michele_Laino
ok!
Michele_Laino
  • Michele_Laino
the work done by the weight force is: 70*9.8*40=...?
calculusxy
  • calculusxy
that's the potential energy right?
Michele_Laino
  • Michele_Laino
yes!
calculusxy
  • calculusxy
27440
Michele_Laino
  • Michele_Laino
the work done against the weight force is equal to the potential energy gained by the elevator
Michele_Laino
  • Michele_Laino
correct!
calculusxy
  • calculusxy
And now I have to add it: 27440 + 1600 = 29040
Michele_Laino
  • Michele_Laino
hint: 70*2*40=5600
calculusxy
  • calculusxy
i got lost after finding the force.
Michele_Laino
  • Michele_Laino
so, we have: 27440+5600=...
calculusxy
  • calculusxy
but why did we need to do 70*2*40?
Michele_Laino
  • Michele_Laino
the work done by the fictitious force, is : m*Ac*h= 70*2*40=...
Michele_Laino
  • Michele_Laino
\[{\text{work = force }} \times {\text{ distance = }}\left( {{\text{m}} \times {{\text{A}}_{\text{c}}}} \right) \times {\text{h}}\]
calculusxy
  • calculusxy
give me a minute...
calculusxy
  • calculusxy
so i do 27440 +5600= 33040
Michele_Laino
  • Michele_Laino
that's right!
calculusxy
  • calculusxy
so that's my answer right?
Michele_Laino
  • Michele_Laino
yes!
calculusxy
  • calculusxy
thank you!!! i have another question may i ask that as well?
Michele_Laino
  • Michele_Laino
ok!

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