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Work done in an accelerating frame.

  • #1

Homework Statement


A man of weight W is in an elevator. The elevator accelerates vertically up at a rate a and at a certain instant has a speed V. The man then climbs a vertical ladder within the elevator at a speed v relative to the elevator. What is the man's rate of expenditure of energy?


Homework Equations


P = Fv


The Attempt at a Solution


So first, I found the apparent weight of the man as
[itex] w=W(1+\frac{a}{g}) [/itex]
which would be the force that he needs to exert in order to continue climbing.
I thought that the power required is just w*v (small v, the speed at which he is climbing in the elevator). But the answer states that it's w*(V+v), the total speed.

I was thinking that the power output should be the same whether the elevator just started to move and when it has already been accelerating for a certain time. So I'm not sure if the answer is wrong, or I've made a conceptual mistake.

Thanks in advance for the help!
 

Answers and Replies

  • #2
verty
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I think you'll get different answers if you treat the elevator as a gravitational system with gravity (a+g), versus treating the man as having speed (v+V) and acceleration (a).

I'd guess that the energy construct is only valid in an inertial frame and takes on a different meaning when changing the gravity. I hope this helps you with your question.
 
  • #3
verty
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Sorry, I should have been more accurate. Energy is valid if one works in an inertial system, but an accelerating system, if treated as a different inertial system, will give different results.
 
  • #4
So to clarify, the power output depends on the frame of reference?

In a frame of reference inside the elevator, the rate of expenditure of energy should still be
[itex] W(1+\frac{a}{g})(v) [/itex]
right? Since the man he has to exert a force of [itex] W(1+\frac{a}{g}) [/itex] and is travelling at speed v. But from an observer who is stationary on the earth, the power is different.
 
  • #5
verty
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198
To clarify, in the elevator frame, his energy is given by:

E_man = E_potential + E_kinetic
= m(a+g)h_elevator + (1/2)mv^2

From the ground, his energy is given by:

E_man = E_potential + E_kinetic
= mg(h_ground + h_elevator) + (1/2)m(v+V)^2

These energy values are not commensurate, so calculating the power will give different answers.
 
  • #6
Ok I get it now :)

Thanks vertigo!
 

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