# Work done in an accelerating frame.

• theloathedone
In summary, the man in the elevator who is accelerating vertically up at a rate a and has a speed V at a certain instant, climbs a vertical ladder within the elevator at a speed v relative to the elevator. The man's rate of expenditure of energy is different depending on the frame of reference, with the elevator frame giving an energy of m(a+g)h_elevator + (1/2)mv^2 and the ground frame giving an energy of mg(h_ground + h_elevator) + (1/2)m(v+V)^2. This leads to different power outputs.

## 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!

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.

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 traveling at speed v. But from an observer who is stationary on the earth, the power is different.

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!

## 1. What is "work done" in an accelerating frame?

"Work done" in an accelerating frame refers to the amount of energy that is transferred or transformed as a result of a force acting on an object within that frame. It is a measure of the change in energy of the object due to the applied force.

## 2. How is work done calculated in an accelerating frame?

The calculation of work done in an accelerating frame involves the force applied to the object and the distance over which the force is applied. Specifically, it is calculated by multiplying the magnitude of the applied force by the displacement of the object in the direction of the force.

## 3. What is the difference between work done in an accelerating frame and work done in an inertial frame?

The main difference between work done in an accelerating frame and work done in an inertial frame is that in an accelerating frame, the object is constantly changing its velocity and therefore its kinetic energy, making the calculation of work done more complex. In an inertial frame, where the object is not experiencing any acceleration, the calculation is simpler and follows the basic formula of work done = force x distance.

## 4. Can work done be negative in an accelerating frame?

Yes, work done can be negative in an accelerating frame. This occurs when the applied force acts in the opposite direction of the displacement of the object. In this case, the force is actually doing work to decrease the object's energy rather than increasing it.

## 5. How does the concept of work done in an accelerating frame relate to Newton's laws of motion?

The concept of work done in an accelerating frame is closely related to Newton's laws of motion, particularly the second law which states that the net force acting on an object is equal to its mass times its acceleration. In an accelerating frame, the net force is constantly changing and therefore the work done is also constantly changing, reflecting the change in the object's kinetic energy.