# Work done on a mass by an ideal machine

• pinokicake
In summary: work done by the ground or support would have to be greater than the work done by the engineer in order for the total energy of the system to be zero.
pinokicake
Hello, I am new. I hope that this is the correct place for this kind of question. I apologize for it being so long or annoying.

## Homework Statement

An engineer that has a mass of 100kg has made an ideal mechanical advantage machine shown in the attachment. Platforms 1 and 2 are attached to the machine. When he steps on platform 1 it goes straight down and platform 2 rises straight up. The maximum weight he can lift using this machine is twice his own. Platform 1 can be lowered to a maximum of 10m. The mechanical advantage of the machine cannot be adjusted.

Assume the mass (m2) is twice the mass of the engineer (m1) and that he gives a slight push down to get the mass moving. Including the push, the work done on the mass as platform 2 rises to its top height of 5m will be equal to:

a) the original potential energy of the engineer
b) the final kinetic energy of the engineer
c) original potential energy of the engineer plus half the final kinetic energy of the engineer
d) original potential energy of the engineer minus the final kinetic energy of the engineer

The answer is C, but I do not understand why my approach does not work and how they reach their conclusion.

## Homework Equations

m1 = mass of the engineer h1 = height of platform 1 v1 = velocity of m1
m2 = mass of the block h2 = height of platform 2 v2 = velocity of m2

Wnc = ΔPE + ΔKE

$$\Delta KE = \frac{1}{2} m \Delta v^2$$

ΔPE= mg(hf-ho)

## The Attempt at a Solution

I said that the work done on the m2 is
W = (1/2m2vf22 - 1/2m2vo22) + (m2gh2f - m2gh2o)

Seeing as that the final velocity would be zero due to it stopping and that its initial height is 0 (0 reference is the ground). The work done on the mass can be simplified to

W = m2gh2f - 1/2m2vo22

Then I said that: m2 = 2m1 h2 =1/2h1 2v2 = v1

How i found the velocities:
Once the engineer pushes the platforms, they move at a constant velocity until they stop. I said that the initial velocity is the moment right after the engineer pushes and the final velocity is when the platforms are at rest. Bellow is how I found the initial velocity of platform 2 in terms of platform 1.

v2 = h2/t
v1=h1/t
h1=2h2
v1 = 2h2/t
v1/2 = h2/t
v1/2 = v2

So this leaves the work done on the mass in terms of the potential and kinetic energy of the engineer as:

W = 2m1g(h1/2) - (1/2)(2m1)(v10/2)2

W = m1gh1 - (1/2)[(1/2)(m1vo12)]

So the work done on the mass is the potential energy from the engineer minus half of the initial kinetic energy of the engineer. This is none of the possible choices and I do not understand were I went wrong. And why my way doesn't work.

4. The solution provided

The net force of both masses is zero so the velocities from the push remain constant. The work done on the mass is W = m2gh2f + 1/2m2vo22, and 2m1=m2 2v2=v1 2h2=h1. The work done on the mass in terms of height, velocity and mass of the engineer is W = m1gh1 + (1/2)[(1/2)(m1vo12)]

5. What I do not understand

How can the final kinetic energy of the engineer not be 0J? The platform is not moving once he moves the maximum distance the platform can travel.

I thought about maybe changing how I define final and initial velocities. I'm thinking maybe I could consider the final velocity to be the velocity right before it stops and the initial velocity would be the platform resting before the push is initiated. If I do that then I get the work done on m2 to be:
W = m2gh2f + 1/2m2vf22.

Once I put it in terms of m1 I get W = m1gh1o + 1/2[(1/2)m1vf12]

If this is the way of getting the answer what was wrong with my original approach to model the problem? Was I not including the push by the engineer?

#### Attachments

• physics forum problem.jpg
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pinokicake said:
How can the final kinetic energy of the engineer not be 0J? The platform is not moving once he moves the maximum distance the platform can travel.

Was I not including the push by the engineer?
The problem does not really specify explicitly whether or not the masses have stopped but let's assume they have. In order for the final KE of the engineer to be 0 something must have caused them to decelerate i.e. work would have needed to be done by the ground or some support. But you don't really need to know this since it was asking about the m2 mass only.

Now the m2 mass would have to be strapped down otherwise it will still keep on moving as well. So this "strap" would also need to do work to decelerate the box to zero. But the problem is only asking about the work done by the machine not the strap (or whatever would cause the deceleration).

The thing I do disagree with the solution that the velocities are constant. I would think that gravity would accelerate the masses otherwise no work has been done on them.

Hello Pino, and welcome to PF.
Your initial apology is quite unnecessary: the platform is there to ask questions and we always claim there are no dumb or annoying questions. In fact you deserve a compliment for such clear and good use of the template and your inventive extension!

I like the exercise because it requires careful reading and some imagination as well. As a physicist I am rather quicly drawn into reading "pulley" or "cogwheel", but never mind.

What I kind of must assume in this story is that the thing is exactly in equilibrium if m2= 2 m1. It is corroborated by the answer, but you are not supposed to need that when confronted with the problem, right? The author also carefully avoids claims that the usual laws of mechanics are violated ("ideal mechanical advantage"). And we read that lowering maximum of 10 m cossreponds to raising maximum 5 m.

The exactly in equilibrium thing means that the forces left and right are balanced. The only external forces are from gravity, so if they cancel the sum of all forces is 0. According to Isaac's law F = dp/dt = ma (with constant mass), we deduce from F=0 that p is constant, hence v is constant.

So if it all initially is hanging still and a push is needed to get going, the work of that push is distributed over the kinetic energy of three times m1. (keep in mind!*)

The exercise doesn't say very much about how the stopping goes, so I propose we answer the question for the "arrival phase", i.e. just before it comes to a stop. Because stopping means doing negative work (to bring the kinetic energy back to zero **). So: until arrival: all the time no F so constant v left and right.

All your work is correct, only the vi and vf assumptions are conflicting with the casus as formulated. If, as you write v1/2 = v2 (which is true at all times!) you cannot have zeroes for the final v of man and load without something that causes that and influences the energy balance (**).

In your point 5 you are thinking in the right direction: final velocities right before it stops ad initial velocities zero. (After all, the push is included in the story -- which carefully avoids mentioning anything initial under a), b), c) or d) ! ). And you come with a way to arrive at the right answer (it definitely is correct!***).

(*) So the work done with the push is two times Δm1v1 ! And the kinetic energies have -- as you show -- a ratio 2:1

(**) This way taking away the speed of the moving load at arrival means doing negative work of 1/2 m1v12 which then leaves W = m2gh2f

Lengthy answer; I think you understand the physics quite well and that's why I try to carefully nudge you to a different interpretation of the exercise formulation. Again: my compliment that you think through and please never ever hesitate to ask if you don't understand (or don't accept) what is out before you !

paisiello2 said:
The thing I do disagree with the solution that the velocities are constant. I would think that gravity would accelerate the masses otherwise no work has been done on them.
Sorry to disagree, paisiello: F = 0 does mean constant velocities (provided mass is constant). Your confusion might be lifted if you realize that gravitational force "does negative work" on the load and positive work on the man.

But I must confess that it looks weird -- to me as well -- if I re-read this last (correct!) sentence.

Alright to disagree, that's what the forum is for anyway.

If F=0 then the work done by the gravitational field on m1 must be canceled out by work done by the platform 1 support reaction. Therefore you must be saying that the machine somehow converts this -ve support reaction work done on m1 to +ve support reaction work done on m2 by platform 2.

I suppose the machine could be a simple lever arm to achieve this but more importantly the problem does not explicitly state this is the case. If the machine were a pulley system then you would have to agree the velocity would not be constant. So I don't know how you can conclude necessarily that F=0.

It doesn't change the final answer either way but I just don't see it being a necessary assumption.

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BvU said:
So if it all initially is hanging still and a push is needed to get going, the work of that push is distributed over the kinetic energy of three times m1. (keep in mind!*)

(*) So the work done with the push is two times Δm1v1 ! And the kinetic energies have -- as you show -- a ratio 2:1

Hello BvU and thank you for your reply/kind words! How is the work with the push 2m1Δv1?

This is how I figured out the work done by the push from the engineer would be:

W = ΔKE + ΔPE

W = [(1/2)m1v1f2 - (1/2)m1v1o2] + (m1gh1f - m1gh1o)

Setting the initial velocity as the velocity of platform 1 before the push is given and the final velocity as the velocity of platform 1 right before it stops simplifies the above equation to:

W = 1/2m1v1f - m1gh1o

(v1o = 0 and m1gh1f=0)

Also isn't the ΔKE for any object (1/2)mΔv2

So how did you get the kinetic energy to be 2m1Δv1?

The thing I do disagree with the solution that the velocities are constant. I would think that gravity would accelerate the masses otherwise no work has been done on them.

I suppose the machine could be a simple lever arm to achieve this but more importantly the problem does not explicitly state this is the case. If the machine were a pulley system then you would have to agree the velocity would not be constant. So I don't know how you can conclude necessarily that F=0.

Hello paisiello2 and thank you for your reply. The work in this problem I think isn't supplied by gravity but the push the engineer provides. So work indeed has been done to platform 1 to get it moving, but it was done for a brief period of time. According to Newton's 2nd law, forces make things accelerate (F=ma). So the platform was initially at rest (and at equilibrium). The engineer pushed for a brief period of time and caused the platform to accelerate down. Once he stops pushing, platform 1 stops accelerating and has a constant velocity (and again is at equilibrium).

In regards to concluding the the system is at equilibrium, in my experience with physics problems usually when they say something like " The maximum weight he can lift using this machine is twice his own." They usually mean when weight he is trying to lift is twice his own, the system is in equilibrium (all forces are equal). This is always been kind of annoying to me because if someone is trying to lift a rock that weighs 3N, they need to pull up with a force that is larger than 3N. So the minimum force they would need to pull (assuming that they are pulling up the rock with their hands and not using a machine) would need to be something that is greater than 3N, meaning that the minimum force required would not be 3N.

But this also brings up an important point that I was wondering about. For a constant force the formula for work is W=F*d*sin(θ). If someone wanted to using only that equation to calculate how much work the engineer did to move platform 1 would they only use the distance the force was applied on? (I don't know if this second question requires a different thread.)

pinokicake said:
Hello paisiello2 and thank you for your reply. The work in this problem I think isn't supplied by gravity but the push the engineer provides. So work indeed has been done to platform 1 to get it moving, but it was done for a brief period of time. According to Newton's 2nd law, forces make things accelerate (F=ma). So the platform was initially at rest (and at equilibrium). The engineer pushed for a brief period of time and caused the platform to accelerate down. Once he stops pushing, platform 1 stops accelerating and has a constant velocity (and again is at equilibrium).
The answer by the OP states otherwise. The definition of work is force x distance so gravity must do work. Any force acting must do work if an object moves in the same direction. If the platform provides a support to resist the weight of mass m1 then it does work in the opposite direction and the net work done on the mass is zero.

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Work of push = Δmv = m1Δv1 + m2Δv2

The plus sign because the machine transforms the downward force on the man platform into an upward force on the load. So both F are in the same directions as the Δ vi

Paisiello: see post #4

BvU said:
Paisiello: see post #4

I think I already responded to that post.

I realize now that the key as to whether or not the velocity is constant is in the original problem definition as pinokicake pointed out:
"The maximum weight he can lift using this machine is twice his own." and "The mechanical advantage of the machine cannot be adjusted."

This implies that the machine is "balanced" so that the weight of m1 will be supported by the weight of m2 and the machine itself. And it would also not matter if the machine is a pulley system or a lever system.

BvU said:
Work of push = Δmv = m1Δv1 + m2Δv2

The plus sign because the machine transforms the downward force on the man platform into an upward force on the load. So both F are in the same directions as the Δ vi

I'm still a little confused. What is seems like you are doing is you are adding the change kinetic energies of both objects (or other wise the work done to both of them). But when I do so, I get a different solution than the one you show.

Here is how I do it:

So the work done on the objects are:

W = ΔKE1 + ΔKE2 (KE1 is the kinetic energy of m1 and KE2 is the kinetic energy of m2)

ΔKE1 = (1/2)m1Δv12 = (1/2)m1vf12 (vo12 = 0)
ΔKE2 = (1/2)m1Δv22 = (1/2)m2vf22(vo22= 0)

W = (1/2)m1vf1 + (1/2)m2vf22

and again seeing as m2=2m1, and vf1=2vf2

W = (1/2)m1vf12 + (1/2)2m1(1/2vf1)2

W = (1/2)m1vf12 + (1/4)m1vf12

W = (3/4)m1vf12

Where is my mistake?

I don't think it is you who is mistaken ! What you calculate is that (after the pushing) the kinetic energy of the mass is half that of the engineer, which is in agreement with C) being the correct answer.

The one who is mistaken is me, in post #8, when I type "Work of push = Δmv = m1Δv1 + m2Δv2". This is plain wrong ! I should have typed Force of push times Δt = Δmv = m1Δv1 + m2Δv2.

Can't say I can convince even myself completely, but with Δv/Δt = a this becomes Force = m1a1 + m2a2.

But I am unhappy to write work = m1a1 s1 + m2a2 s2 to match with the 3/4 (because s2 = s1/2). I could just as well have written that the F = 2 m1a1 has to be exercised over the full s1.

So, for a change, I am stuck and need someone else to point out the flaw in this!

I think I figured out a good way of conceptually looking at this problem. The energy that is contained in the push has to do 2 things. It has to get platform 1 moving and it has to get platform 2 moving. So the energy that is contained in the push is divided in half. The the energy that is put into m2 comes from half of the energy of the push (which is equal to the kinetic energy of m1) and the potential energy that was in m1 initially.

Is there any flaw in this kind of thinking?

I don't think that is a good way of conceptualizing it. The kinetic energy in the push is just that. It doesn't have to do anything to get platform 2 moving.

The ideal machine however converts this kinetic energy directly to the platform mass giving it kinetic energy which simply because of the values given turns out to be half.

Post 13: no. Not a good way. And yes: flawed.

Post 14: That doesn't help. There is no kinetic energy in a push. And the push definitely must get m2 moving too.

There is no conversion by the machine. At best it transfers a force. And "turns out to be half" is cryptic.

----------------------------------

back to post 13:

Yes, there is (a flaw). The work that the engineer does in pushing is converted into kinetic energy.
Two thirds go into his own (+platform) kinetic energy:

1/2 m1v12

and one third goes into the kinetic energy of the load (+platform):

1/2 m2v22 = 1/2 * 2 m1 * (v1/2)2 = 1/4 m1v12 .​

That 's the way it stays until arrival. At arrival the platform has total energy = original potential energy of the engineer plus half the final kinetic energy of the engineer, conform answer c.

--------------------------

Now my own stupidities, that may have contributed to the confusion: in post 12 I correct my own post 9. Post 9 is wrong, but post 12 isn't correct either!

Force of push times Δt = Δmv = m1Δv1 + m2Δv2 is wrong. Force doesn't act directly on platform 2 & mass 2.
This is F = ma in an incorrect interpretation. Thing to do is separate into FBDs for m1 and m2 and consider the force the platform executes on the engineer during pushing too (up = positive, g = 9.81 m/s2):

on m1

-F + F2→1 - m1 g = m1 a1 ##\ \ ## so ##\ \ ## F2→1 = F + m1 g + m1 a1
on m2

F1→2 - m2 g = m2 a2

With a2 = - 1/2 a1 m2 = 2 m1 and

F2→1 = 1/2 F1→2##\ \ ## so ##\ \ ## 2 F2→1 = F1→2
this becomes
2F + 2m1 g + 2m1 a1 - m2 g = m2 a2 ##\ \ ## or ##\ \ ## 2F + 2m1 g + 2m1 a1 - 2m1 g = 2m1 (-a1/2)​
so that
F + m1 a1 = m1 (-a1/2) ##\ \ ## ⇔ ##\ \ ## F = - 3/2 m1 a1
Must admit that at first I was a little surprised at the 3/2. (previously I thought 2, so 4/2). But the 1 is used to accelerate m1 with an acceleration a1 and 1/2 is "multiplied by 2 by the machine" to accelerate 2m1 with an acceleration a1/2.

Someone can shoot a hole in this ? Receives a hearty thank you from me, but I don't think there are any takers...By the way: pino thanks/compliments to your teacher for bringing in such a simple yet challenging exercise!

BvU said:
Post 14: That doesn't help. There is no kinetic energy in a push. And the push definitely must get m2 moving too.

There is no conversion by the machine. At best it transfers a force. And "turns out to be half" is cryptic.
Suppose I should have said "from a push" instead of "in a push" but that seems like splitting hairs. So does the semantics between "convert" and "transfer".

In the context of the question I don't think the push has to do anything other than give KE to the mass on platform 1. It's the machine that is converting or transferring this energy from -ve work done on m1 into +ve work done on m2. If the machine wasn't there then none of this would happen.

And recall that the answer to the OP involved 1/2 of the KE of mass m1. So I don't feel I was bring cryptic in referring to this.

BvU said:
Someone can shoot a hole in this ? Receives a hearty thank you from me, but I don't think there are any !

I'll give it a try!

What does all your formula working have to do with answering the question about work done on mass m2?

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See 9 and 12. Truthfinding out of a sense of wonder why it wasn't as I supposed.

OK but it might be confusing to the OP since he and I were talking about the reasons why the KE portion is half that of the engineer.

Last edited:
KE is not a mass.

You're right, but a "mass" in the most general sense can have KE. So I was referring to the mass of the engineer that has KE.

I wanted to point out that F = - 3/2 m1 a1 is in agreement with W = 3/2(KE)

The proof:

W = ΔKE1 + ΔKE2

ΔKE1 = (1/2m1v1f2 - 1/2m1vo12) = 1/2m1v1f2

ΔKE2 = (1/2m2v2f2 - 1/2m2vo22) = 1/2m2v2f2

W = 1/2m1v1f2 + 1/2m2v2f2

m2 = 2m1 and v1 = (1/2)v2

W = 1/2m1v1f2 + 1/2(1/2m1v1f2)

W = KE1 + 1/2 KE1 → W = 3/2 KE1 or W = (3/2) (1/2m1v1f2)

Vf/2 = a*h (h is the distance, this is from the kinematics equation vf2 = v02 + 2ax)

W = 3/2 m*a*h → W = 3/2F*h, and the negative is not included because the force is in the same direction as the movement of platform 1.

And also pointing out another flaw in my original thinking of how the energy is transferred. If truly half of the energy that was stored in his body went into moving the mass, then the work done on m2 would have been the potential energy of the engineer plus this kinetic energy rather than half his kinetic energy.

I still can't think up of an intuitive way of looking at the problem and thinking "Oh yeah, if the mass is the maximum weight the engineer can move than the work done on the mass would be all of the potential energy of the engineer and half of his kinetic energy. And if the mass was less than the maximum weight the engineer could move, it would be all of his potential energy minus his kinetic energy." I will keep chewing on this.

## 1. What is "work done on a mass by an ideal machine"?

Work done on a mass by an ideal machine refers to the amount of energy transferred to a mass by a machine that is considered to have no energy losses due to friction, heat, or other factors. This type of machine is often used as a theoretical model for calculating work in scientific and engineering calculations.

## 2. How is work done on a mass by an ideal machine calculated?

The work done on a mass by an ideal machine can be calculated by multiplying the force applied to the mass by the distance the mass moves in the direction of the force. This is represented by the equation W = Fd, where W is the work done, F is the applied force, and d is the distance moved.

## 3. What is the significance of an ideal machine in calculating work?

An ideal machine is used in calculations to simplify the process by eliminating any energy losses due to friction. This allows for more accurate calculations and theoretical models to be created. In real-world scenarios, machines will always have some level of energy loss, so using an ideal machine allows for a more precise understanding of the principles behind work and energy.

## 4. Can an ideal machine exist in reality?

No, an ideal machine is a theoretical concept and cannot exist in reality. In order for a machine to perform work, there will always be some level of energy loss due to factors such as friction, heat, and other inefficiencies. However, ideal machines are useful for understanding the fundamental principles of work in a simplified manner.

## 5. What are some examples of ideal machines?

Some examples of ideal machines include frictionless pulleys, massless springs, and perfectly smooth surfaces. These machines are used in theoretical models and calculations, but do not exist in reality. In real-world scenarios, these ideal machines can be approximated, but will always have some level of energy loss.

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