# Work done in an accelerating frame.

## 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?

P = Fv

## The Attempt at a Solution

So first, I found the apparent weight of the man as
$w=W(1+\frac{a}{g})$
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!

verty
Homework Helper
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.

verty
Homework Helper
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.

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
$W(1+\frac{a}{g})(v)$
right? Since the man he has to exert a force of $W(1+\frac{a}{g})$ and is travelling at speed v. But from an observer who is stationary on the earth, the power is different.

verty
Homework Helper
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.

Ok I get it now :)

Thanks vertigo!