Climbing a ladder in an elevator

[KungFu]

Homework Statement

A man with mass m_1 is in an elevator with mass 2m_1. The elevator accelerates vertically up at a rate a and at a certain time t has a speed V . The man climbs a vertical ladder within the elevator at a speed v relative to the elevator. What is the man's power output?

Relevant Equations

Newtons second law for finding the apparent weight F of the man

and then

finding the power output P = apparent weight times the speed

I have found the apparent weight of the man in the accelerating elevator

My question is about what speed I should use in the formula for the power output P = F ⋅ speed

My common sense tells me that the speed I should us is the speed v the man is climbing the ladder , because if the climbingspeed is zero then the man is standing still on the latter and is therefore not spending any energy.
However, the answer in the book is using the total speed, that is v + V , the sum of the climbingspeed and the elevator speed. Why ?

 

[haruspex]

Homework Statement:: A man with mass m_1 is in an elevator with mass 2m_1. The elevator accelerates vertically up at a rate a and at a certain time t has a speed V . The man climbs a vertical ladder within the elevator at a speed v relative to the elevator. What is the man's power output?
Relevant Equations:: Newtons second law for finding the apparent weight F of the man

and then

finding the power output P = apparent weight times the speed

I have found the apparent weight of the man in the accelerating elevator

My question is about what speed I should use in the formula for the power output P = F ⋅ speed

My common sense tells me that the speed I should us is the speed v the man is climbing the ladder , because if the climbingspeed is zero then the man is standing still on the latter and is therefore not spending any energy.
However, the answer in the book is using the total speed, that is v + V , the sum of the climbingspeed and the elevator speed. Why ?

It is not completely clear what final answers you and the book have. Do you get m₁(a+g)v? And the book has the same but with v+V?

 

[etotheipi]

I think this question is more complicated than it seems. Let us, initially, abstract away the details of "the man" as something that has chemical energy available, and imagine he is a particle being pulled upward by a force $\vec{F}$.

Suppose we want to calculate the effective power of this force. Since the velocity vector is not frame invariant the power of this force would not be frame invariant. So the power of this force in the elevator frame uses the speed $v$, but the power of this force in the inertial ground frame uses the speed $v+V$. (For instance, things like work and kinetic energy are frame variant, but so long as you work in a single frame of reference throughout energy conservation still applies)

But this simplified scenario is not so interesting. Now we consider a man who is doing a pull-up on a fixed rung in the elevator (and his centre of mass is rising at a constant speed $v$ in the elevator frame).

In the elevator frame, his hands are fixed so the contact force has no power. His effective weight acts through his centre of mass, though, so this force as a negative power. But since his kinetic energy in the elevator frame is constant, he must be providing a chemical power $P_{chem}$ that equals $m(a+g)v$.

Now we consider the lab frame. If the contact force has magnitude $F$ then we have that $F = m(a+g)$. This means that the power of the contact force is $FV$, since the rung in the elevator is rising at $V$, and the power of his weight is $-mg(V+v)$, since his centre of mass is rising at $V+v$. Now let us consider conservation of energy, $$P_{contact} + P_{weight} + P_{chem} = \frac{dT}{dt}$$ $$m(a+g)V - mg(V+v) + P_{chem} = \frac{dT}{dt} = \frac{d}{dt} \left(\frac{1}{2}m(V+v)^2 \right) = m(V+v)a$$ $$maV - mgv + P_{chem} = maV + mav$$ and this gives us $$P_{chem} = m(a+g)v$$ which is, as expected, what we got last time around!

So we conclude that the chemical power of the man is indeed frame invariant (which makes sense, since everyone agrees on how much food he ate beforehand!). But I think that the solution manual has it wrong, the speed should be $v$ if we are interested in chemical power.

 

[Halc]

My common sense tells me that the speed I should us is the speed v the man is climbing the ladder , because if the climbingspeed is zero then the man is standing still on the latter and is therefore not spending any energy.
However, the answer in the book is using the total speed, that is v + V , the sum of the climbingspeed and the elevator speed. Why ?

Your common sense seems right. Speed V is relevant to the power expended by whatever is accelerating that elevator, but the man is not doing this. He exerts continuous power at all times as he climbs the ladder against a fixed weight, so V and t (and mass of the elevator and of the ladder) need not be taken into consideration.

 

[PeroK]

Homework Statement:: A man with mass m_1 is in an elevator with mass 2m_1. The elevator accelerates vertically up at a rate a and at a certain time t has a speed V .

However, the answer in the book is using the total speed, that is v + V , the sum of the climbingspeed and the elevator speed. Why ?

The speed $V$ must be specified relative to some inertial frame of reference: the ground frame, probably. By the principle of relativity, we can change to any other inertial reference frame and take the speed $V'$ in that frame to be $0$ (or anything else). The speed $V$ should not appear in the answer.

Note that $V$ is relevant to the engine (at rest relative to the Earth) and whose power consumption must be increasing with $V$. Rather than the man's power consumption increasing with $V$.

 

[PeroK]

PS Note that we can also consider $a+g$ as the "effective" gravity and the problem is equivalent to climbing a ladder under this effective gravitational force. And instead of an accelerating elevator on Earth, we have a room at rest on the surface of a planet with gravity $a + g$.

That nicely applies both the principle of relativity and the equivalence of accelerated reference frames and gravity.

 

[KungFu]

I think this question is more complicated than it seems. Let us, initially, abstract away the details of "the man" as something that has chemical energy available, and imagine he is a particle being pulled upward by a force $\vec{F}$.

Suppose we want to calculate the effective power of this force. Since the velocity vector is not frame invariant the power of this force would not be frame invariant. So the power of this force in the elevator frame uses the speed $v$, but the power of this force in the inertial ground frame uses the speed $v+V$. (For instance, things like work and kinetic energy are frame variant, but so long as you work in a single frame of reference throughout energy conservation still applies)

But this simplified scenario is not so interesting. Now we consider a man who is doing a pull-up on a fixed rung in the elevator (and his centre of mass is rising at a constant speed $v$ in the elevator frame).

In the elevator frame, his hands are fixed so the contact force has no power. His effective weight acts through his centre of mass, though, so this force as a negative power. But since his kinetic energy in the elevator frame is constant, he must be providing a chemical power $P_{chem}$ that equals $m(a+g)v$.

Now we consider the lab frame. If the contact force has magnitude $F$ then we have that $F = m(a+g)$. This means that the power of the contact force is $FV$, since the rung in the elevator is rising at $V$, and the power of his weight is $-mg(V+v)$, since his centre of mass is rising at $V+v$. Now let us consider conservation of energy, $$P_{contact} + P_{weight} + P_{chem} = \frac{dT}{dt}$$ $$m(a+g)V - mg(V+v) + P_{chem} = \frac{dT}{dt} = \frac{d}{dt} \left(\frac{1}{2}m(V+v)^2 \right) = m(V+v)a$$ $$maV - mgv + P_{chem} = maV + mav$$ and this gives us $$P_{chem} = m(a+g)v$$ which is, as expected, what we got last time around!

So we conclude that the chemical power of the man is indeed frame invariant (which makes sense, since everyone agrees on how much food he ate beforehand!). But I think that the solution manual has it wrong, the speed should be $v$ if we are interested in chemical power.

Thanks for the detailed answer mate. So, to be sure I get this correct, when you work in the elevator frame, and using the conservation of energy in the same way as you did in the lab frame, you get:

$$P_{contact} + P_{weight} + P_{chem} = \frac{dT}{dt}$$
$$P_{contact}=0 , P_{weight} = - m(a+g)v$$ and $$\frac{dT}{dt} = 0$$
so, $$P_{chem} = m(a+g)v $$

 

[KungFu]

It is not completely clear what final answers you and the book have. Do you get m₁(a+g)v? And the book has the same but with v+V?

yes, that is correct , I wonder if you can get the answer in the book if you just view the problem in a different way? I am not sure how though. I fond this thread discussing the same problem, but the answers did not made it clear for me : https://www.physicsforums.com/threads/work-done-in-an-accelerating-frame.470263/

 

[PeroK]

yes, that is correct , I wonder if you can get the answer in the book if you just view the problem in a different way? I am not sure how though. I fond this thread discussing the same problem, but the answers did not made it clear for me : https://www.physicsforums.com/threads/work-done-in-an-accelerating-frame.470263/

The answer in the book is wrong. See post #6.

 

[KungFu]

The answer in the book is wrong. See post #6.

ok, cool, that makes sense , thanks !

 

[etotheipi]

$$P_{contact} + P_{weight} + P_{chem} = \frac{dT}{dt}$$ $$P_{contact}=0 , P_{weight} = - m(a+g)v$$ and $$\frac{dT}{dt} = 0$$ so, $$P_{chem} = m(a+g)v $$

Yeah that's right, the chemical power is just the rate at which internal chemical energy is being converted into mechanical energy (so it differs slightly from the other two powers, which arise due to external interactions)

We call the total energy of the system (the man) the sum of the chemical energy and kinetic energy, $E = T + E_{chem}$, and equate the net power across the boundary by any means of transfer of energy to the rate of increase of the energy of that system $$P_{contact} + P_{weight} = \frac{dE}{dt} = \frac{dT}{dt} + \frac{dE_{chem}}{dt}$$ and we then just transpose the $\frac{dE_{chem}}{dt}$ to the LHS and set $P_{chem} = -\frac{dE_{chem}}{dt}$. That's how you can think of setting up the equation.

 

[KungFu]

Yeah that's right, the chemical power is just the rate at which internal chemical energy is being converted into mechanical energy (so it differs slightly from the other two powers, which arise due to external interactions)

We call the total energy of the system (the man) the sum of the chemical energy and kinetic energy, $E = T + E_{chem}$, and equate the net power across the boundary by any means of transfer of energy to the rate of increase of the energy of that system $$P_{contact} + P_{weight} = \frac{dE}{dt} = \frac{dT}{dt} + \frac{dE_{chem}}{dt}$$ and we then just transpose the $\frac{dE_{chem}}{dt}$ to the LHS and set $P_{chem} = -\frac{dE_{chem}}{dt}$. That's how you can think of setting up the equation.

ok, but what about the potential energy? I can see that the expression m(a + g)v equals the derivative of the potential energy $$\frac{d(m(a + g)h)}{dt} = m(a+g)v$$ , the work done by a force lifting a mass a certain vertical distance is given by the change in potential energy , so from this perspective it also makes sense to take the derivative of the potential energy with respect to the distance to obtain the expression for the chemical energy. I am just trying to understand if really the chemical energy is the same as the gravitational potential energy.

 

[etotheipi]

There is an equivalence between gravitational work and gravitational potential energy, the former equals the negative of the change of the latter i.e. $dU_{grav} = - m\vec{g} \cdot d\vec{x}$, or you will usually see this written in a form something like $m\vec{g} = -\nabla U_{grav}$ (or in one dimension, $mg_x = -\frac{dU}{dx}$).

We could indeed substitute $P_{weight} = \vec{F}_{grav} \cdot \vec{v}_{CM} = -\frac{dU_{grav}}{dt}$ if you would prefer to speak in terms of gravitational potential energy.

It doesn't make sense to speak about the spatial derivative of the chemical energy, instead just think of chemical energy as a form of energy internal to the system. Perhaps on a microscopic level you could attempt to decompose this energy in terms of interactions between molecules and things but that is not at all relevant here.