# Work done in lifting and lowering an object

• B
• rudransh verma
In summary, the energy of a system is constant if the forces that apply it change, but if the forces are the same the system will still move.
rudransh verma
Gold Member
##\Delta K=K_f-K_i=W_a+W_g##.(##W_a##, work done by applied force and ##W_g##, work done by gravity) In case of uniform motion with velocity u, kinetic energy is equal. Change is zero. ##W_a=-W_g## If one force transfers energy into the system then the other takes out of the system. Energy of the system remains constant. Body stays in constant motion.
But when we talk about lifting a book from rest and putting it on shelf, one force provides energy to the system and other one takes out. The change in kinetic energy is zero. Then how does the body move(accelerate) from rest to the shelf.( It’s kinetic energy is always zero)?

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rudransh verma said:
It’s kinetic energy is always zero
No it isn't. But it can be exchanged for potential energy at arrival.
And it can be chosen as close to zero as desired in the process of displacement.

##\ ##

BvU said:
No it isn't. But it can be exchanged for potential energy at arrival.
And it can be chosen as close to zero as desired in the process of displacement.

##\ ##
What are you trying to say that ##W_a## is not equal to ##W_g##?

Your question is not well defined. You specified "uniform motion with velocity u" and then asked about how it accelerates. If it's moving uniformly, it doesn't accelerate.

Ibix said:
You specified "uniform motion with velocity u" and then asked about how it accelerates. If it's moving uniformly, it doesn't accelerate.
No there are two parts here. One when the body is moving with constant u and the other is when we take a book from floor and put it on shelf.

I see.

For a book moving from one shelf to another, the kinetic energy is initially zero. It becomes non zero. Then it becomes zero again. Which forces are responsible for the changes in velocity?

Ibix said:
I see.

For a book moving from one shelf to another, the kinetic energy is initially zero. It becomes non zero. Then it becomes zero again. Which forces are responsible for the changes in velocity?
Applied force>weight, body accelerates
then applied force =weight , body cruise,
then Weight>applied force, body deaccelerates, then weight=applied force, body stops.
I think first the kinetic energy is zero, increases becomes constant , decreases and then again becomes zero.

Right. So the kinetic energy isn't always zero, although it is zero at beginning and end of the journey. Does that answer your question?

Ibix said:
Right. So the kinetic energy isn't always zero, although it is zero at beginning and end of the journey. Does that answer your question?
So first ##K_f-K_i=0 => W_a=-W_g=0##
##K_f-K_i>0##, ##W_a>W_g## energy is given to system, body accelerates
##K_f-K_i=0##, ##W_a=-W_g##, body cruise
##K_f-K_i<0##, ##W_a<W_g##, body loses energy, deacelerates.
##K_f-K_i=0##, ##W_a=-W_g=0##, body stops

I'd say your notation is terrible, because what you are calling ##K_f## in one line is what you are calling ##K_i## in the next and the ##W_a## and ##W_g## values in each line are different. But I think you have the idea. In the first phase your muscles (or whatever is lifting the book) produce a force greater than ##mg##, and the body accelerates, you doing more positive work than gravity does negative work. Then you reduce the force to ##mg## and the forces balance and the energy inflow from your muscles matches the outflow from gravity. Finally you reduce your force (and, hence, energy inflow) while the gravitational force (and hence energy outflow) remains constant, so the book decelerates.

rudransh verma said:
Then how does the body move(accelerate) from rest to the shelf.( It’s kinetic energy is always zero constant)?
It doesn't need to accelerate. In the model being used it's already in motion when it leaves the floor and is still in motion when it arrives at the shelf.

Ibix said:
I'd say your notation is terrible, because what you are calling ##K_f## in one line is what you are calling ##K_i## in the next and the ##W_a## and ##W_g## values in each line are different. But I think you have the idea. In the first phase your muscles (or whatever is lifting the book) produce a force greater than ##mg##, and the body accelerates, you doing more positive work than gravity does negative work. Then you reduce the force to ##mg## and the forces balance and the energy inflow from your muscles matches the outflow from gravity. Finally you reduce your force (and, hence, energy inflow) while the gravitational force (and hence energy outflow) remains constant, so the book decelerates.
##W_a(0)+W_g(0)+W_a(+)+W_g(-)+W_a+W_g+W_a(-)+W_g(+)+W_a(0)+W_g(0)=\Delta k##
Initially and finally both ##W_a## and ##W_g## are zero. In accelerating and de accelerating the work done by the applied forces are equal and opposite. Similarly for gravity. So they cancel out. Only works left are the ##W_a=W_g##.
So in both lifting and cruising cases. ##\Delta K=W_a+W_g##. In both cases initial and final velocity are zero, so ##W_a=-W_g=-mgd\cos\theta##.
It does not depend on applied force just the weight mg
Right?

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rudransh verma said:
Right?
I have no idea. You haven't bothered to define any of the quantities you are using.

russ_watters and BvU
Ibix said:
I have no idea. You haven't bothered to define any of the quantities you are using.
##W_a(0)## is simply work done by applied force which is zero.
##W_g(+)## is simply work done by gravity which is some +ve value.
And like that, all of them.

That's still useless. Let's set this out sensibly.

The book is initially at rest on one shelf and is moved to another shelf in a motion consisting of three phases. In the first phase a force ##F##, which is greater than the book's weight, is applied to accelerate the book until it reaches a speed ##u##. In the second phase the force is instantaneously reduced to equal the book's weight and it moves upwards at constant speed ##u##. In the third phase, the force is instantaneously reduced to ##F'##, which is less than the book's weight, until it comes to rest on the other shelf.

During phase 1, the force ##F## does work ##W_{a1}## and gravity does work ##W_{g1}##. Because the book is accelerating upwards, ##W_{a1}>0## and ##W_{g1}<0##, and the total work done on the book is ##W_1=W_{a1}+W_{g1}##, which is positive. Since its initial kinetic energy was zero, its final kinetic energy is ##K_1=0+W_1##.

During phase 2, the force does work ##W_{a2}## and gravity does work ##W_{g2}##. Again the signs are opposite but, since the forces have equal magnitudes, ##W_{a2}=-W_{g2}##. Thus no net work is done on the book and its kinetic energy at all times during this phase is ##K_1##.

During phase 3, the force ##F'## does work ##W_{a3}## and gravity does work ##W_{g3}##. Once again the signs are opposite with ##W_{a3}>0## and ##W_{g3}<0##. However, since the magnitude of the force is less than the weight, this time the net work done on the book, ##W_3=W_{a3}+W_{g3}##, is negative. At the end of this phase the book is at rest, so has zero kinetic energy. Thus we require the kinetic energy at the end of this phase, ##K_3##, be zero. So ##K_3=0=K_1+W_3##, which implies that ##W_3=-W_1##.

Do you see how I've defined everything clearly here? I started with an overview of the experiment, explaining what I was doing and that it could be broken into three phases. Then I covered each phase in detail, defining each term as I went, relating it to the motions and the forces. You should be able to see where every term comes from, how it's been defined, and what part of the motion it relates to. That's how to lay out a description of an experiment so other people can understand it - we cannot read your mind, so you need to write everything down.

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Lnewqban and rudransh verma
Ibix said:
During phase 2, the force does work Wa2 and gravity does work Wg2. Again the signs are opposite but, since the forces have equal magnitudes, Wa2=−Wg2. Thus no net work is done on the book and its kinetic energy at all times during this phase is K1.
And That is what the book wrote directly ##\Delta K=K_f-K_i=W_{a2}+W_{g2}## for the non cruising case.
It should be ##\Delta K=W_1+W_{a2}+W_{g2}+W_3## This tells us that ##W_{a2}=-W_{g2}##. Work done by the applied force doesn’t depend on the applied force, just ##-mgd\cos\theta##.
Is this right? Because in cruising case also work done by the applied force does depend on applied force. In non cruising case work is done by the applied force in accelerating and non accelerating part. I don’t get it!

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During all of the motion work is done on the book by the force, but work is also done with the opposite sign by gravity. So the book first gains kinetic energy as the large initial force does more positive work than gravity does negative work. Then it loses kinetic energy as the force reduces and gravity is doing more negative work than the force is doing positive work. The end result is that the total work done on the book is zero - the energy used ends up as gravitational potential, which can be recovered by dropping the book off the shelf.

It's like me giving you £10 and then you giving that £10 to someone else. At the end of the day you have no more money than you started with (##\Delta\mathrm{Money}=0##), but for a while you had £10. The force transfers energy to the book and the book transfers it to the gravitational potential, ending up with no more kinetic energy than it started with.

Lnewqban
Ibix said:
During all of the motion work is done on the book by the force, but work is also done with the opposite sign by gravity. So the book first gains kinetic energy as the large initial force does more positive work than gravity does negative work. Then it loses kinetic energy as the force reduces and gravity is doing more negative work than the force is doing positive work. The end result is that the total work done on the book is zero - the energy used ends up as gravitational potential, which can be recovered by dropping the book off the shelf.

It's like me giving you £10 and then you giving that £10 to someone else. At the end of the day you have no more money than you started with (##\Delta\mathrm{Money}=0##), but for a while you had £10. The force transfers energy to the book and the book transfers it to the gravitational potential, ending up with no more kinetic energy than it started with.
That is all right . I got that. What I don’t get is
rudransh verma said:
Work done by the applied force doesn’t depend on the applied force, just −mgdcos⁡θ.
Is this right? Because in cruising case also work done by the applied force does depend on applied force. In non cruising case work is done by the applied force in accelerating and non accelerating part. I don’t get it!
Even though the value of ##W_a## is ##-mgd\cos\theta## but this work does depend on applied force. If there is no applied force there is no work by the applied force. How will the body move up without applied force?

rudransh verma said:
How will the body move up without applied force?
You mean, if you throw a ball straight upwards why does it keep going upwards?

In terms of energy, because it has kinetic energy which is converted to potential energy. If its initial speed is ##u## then its initial kinetic energy is ##mu^2/2##. When it has risen a height ##h## that has reduced to ##mu^2/2-mgh##. It stops rising when that quantity is zero.

Ibix said:
You mean, if you throw a ball straight upwards why does it keep going upwards?
No! In resnik it’s proven that when the initial and final Kinetic energy are zero or equal like in a lift of a book to a shelf then ##W_a=-W_g=-mgd\cos\theta##(##W_a## is work done by applied force and ##W_g## is work done by gravity) which means the work done by the applied force does not depend on the force applied. Is this right to say?

rudransh verma said:
No! In resnik it’s proven that when the initial and final Kinetic energy are zero or equal like in a lift of a book to a shelf then ##W_a=-W_g=-mgd\cos\theta##(##W_a## is work done by applied force and ##W_g## is work done by gravity) which means the work done by the applied force does not depend on the force applied. Is this right to say?
If one is given that the diagonal distance ##d## between two shelves is fixed and if one is given the angle ##\theta## between the two shelves and if one is given the [uniform] local acceleration of gravity ##g## and if one is given that initial and final kinetic energy are zero and if one is given that no other forces act then any pattern of applied force that is compatible with the givens will result in the same value for ##W_a##.

For an arbitrary force, the work done depends on the force applied.

For a force profile that is is bound by the givens of a problem, it may turn out that the work done is fixed and knowable even when the exact force profile is not completely constrained.

As anyone who has lifted a book from one shelf to another can attest. It is possible to do it quickly along one path or slowly along another. The force profiles will be different. The initial and final energies will be identical.

Lord Jestocost
jbriggs444 said:
or a force profile that is is bound by the givens of a problem, it may turn out that the work done is fixed and knowable even when the exact force profile is not completely constrained.
But to say the work done does not depend on applied force is actually not right. It will be fixed that is true.
Even though in the formula it’s ##-mgd\cos\theta## and no applied force but it’s not right.

rudransh verma said:
But to say the work done does not depend on applied force is actually not right. It will be fixed that is true.
Even though in the formula it’s ##-mgd\cos\theta## and no applied force but it’s not right.
Don't be silly.

rudransh verma
With that, I am afraid that I will have to censor myself from any future involvement in your threads.

hutchphd and BvU
rudransh verma said:
the work done by the applied force does not depend on the force applied. Is this right to say?
The work done by a force depends during a movement is equal to the change in kinetic and potential energy and does not depend on the magnitude of the force if that change is held constant.

So consider moving a book from one shelf to another higher one. The book gains zero kinetic energy (it is at rest at the start and end of the motion) and the change in potential energy depends solely on the height difference. You can use a very high force for a very short distance and throw the book on to the shelf. Or you can use a force barely stronger than the weight and lift the book smoothly. Either way, the work done by the force is the same.

Lnewqban
jbriggs444 said:
With that, I am afraid that I will have to censor myself from any future involvement in your threads.
No bro! Don’t do that. We should always help each other. I did not mean any disrespect.
Please explain why you think I am going wrong.

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Ibix said:
The work done by a force depends during a movement is equal to the change in kinetic and potential energy and does not depend on the magnitude of the force if that change is held constant.

So consider moving a book from one shelf to another higher one. The book gains zero kinetic energy (it is at rest at the start and end of the motion) and the change in potential energy depends solely on the height difference. You can use a very high force for a very short distance and throw the book on to the shelf. Or you can use a force barely stronger than the weight and lift the book smoothly. Either way, the work done by the force is the same.
Ok! If ##\Delta K = W=F.d## and left side is constant then right side is too. Work done doesn’t depend on force. Any amount of force will produce the same work.
But there is a confusion here in ##W=Fd## , d is the distance up to which the force is applied and up to which the body travels. It’s not a momentary push.
I have added the paragraph below and I think the condition ##W_a=-W_g## is when the body cruise, not before and after when the body accelerates or deaccelerates. Only then it’s ##-mgd\cos\theta## and the value of work by applied force is either ##mgd## or ##-mgd##. But it does require that the Kinetic energy remains unchanged or becomes zero before and after the displacement. And during the phase of cruising the two forces will be equal to mg. ##F_a## and ##F_g##. But it doesn’t mean that work done by applied force doesn’t depend on force applied. It just that the applied force does not vary.

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I find that last paragraph a little confusing.
The lifting force is whatever force is necessary to overcome gravity, it can't be less.
If more, then the object is accelerated, as it happens at the beginning (upwards net force and acceleration) and end (downwards net force and acceleration) of the displacement of the object.

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Lnewqban said:
I find that last paragraph a little confusing.
The lifting force is whatever force is necessary to overcome gravity, it can't be less.
If more, then the object is accelerated, as it happens at the beginning (upwards net force and acceleration) and end (downwards net force and acceleration) of the displacement of the object.
Equations 7-16 and 7-17 apply to any situation in which an object is lifted or
lowered, with the object stationary before and after the lift.
They are independent
of the magnitude of the force used. For example, if you lift a mug from the floor

The first line is very confusing in bold. It is not clear whether the eqns 7-16 and 7-17 are applicable during cruising or the entire motion of the lift.

Using the example of the elevator of a previous thread:
One person is transported from the first to the second floor.
The elevator was not moving the body of the person, or providing kinetic energy to it, at the beginning or end of its stroke.

Once at the second floor, the body of the person had additional mechanical energy, which could only be measured by the gained altitude or distance traveled against the natural gravity acceleration or weight force.
An external observer can't tell how fast the elevator moved up or how the tension in the cable varied during the first to second floor stroke.

The energy that the elevator transferred to the person's body was not natural, fuel had to be combusted to generate mechanical work, then electrical work (motor of the elevator), then mechanical work again (variable tension of the cable times height of second floor respect to first floor).

Now for the bold text:
The tension of the cable must have ben variable during the stroke, since the person was accelerated first, moved at constant velocity by mid stoke and finally decelerated to a stop before the doors of the elevator opened at the second floor.

That variation of the force could have taken many forms.
It could have cycled several times during the stroke, or it could have been a brutal initial force followed by a gradual reduction, or vice-verse; or it could have been very close to the magnitude of the weight force, slowing the stroke very much.

Those equations only describe initial and final conditions, disregarding the process in between.
If the person of our example is returned back down to the first floor, his/her body has not gained any mechanical energy.

https://courses.lumenlearning.com/b.../potential-energy-and-conservation-of-energy/

:)

Lnewqban said:
The lifting force is whatever force is necessary to overcome gravity, it can't be less.
If the object is moving upward and slowing down, the lifting force is less than ##mg##.

rudransh verma
rudransh verma said:
But there is a confusion here in W=Fd , d is the distance up to which the force is applied and up to which the body travels. It’s not a momentary push.
What's a "momentary push"? It has to last for some length of time, and during that time the object moves a distance ##d##.

Mister T said:
What's a "momentary push"? It has to last for some length of time, and during that time the object moves a distance ##d##.
Oh! It’s the same thing.

Ibix said:
You can use a very high force for a very short distance and throw the book on to the shelf. Or you can use a force barely stronger than the weight and lift the book smoothly. Either way, the work done by the force is the same.
When using high forces for very short distance mean momentary push. In my book resnik they are not using this thing but constant force throughout the motion and displacement. Even though after a set time both will produce same change in energy.

rudransh verma said:
In my book resnik they are not using this thing but constant force throughout the motion and displacement.
If the book starts and stops, they cannot be using constant force.

Lnewqban

## 1. What is work done in lifting and lowering an object?

The work done in lifting and lowering an object is the amount of energy required to move an object from one position to another against the force of gravity.

## 2. How is work calculated in lifting and lowering an object?

The work done in lifting and lowering an object is calculated by multiplying the force applied to the object by the distance it is moved in the direction of the force.

## 3. Does the weight of the object affect the work done in lifting and lowering it?

Yes, the weight of the object does affect the work done in lifting and lowering it. The greater the weight of the object, the more work is required to move it against the force of gravity.

## 4. Is the work done in lifting and lowering an object the same regardless of the speed at which it is moved?

No, the work done in lifting and lowering an object is not the same regardless of the speed at which it is moved. The work done is directly proportional to the speed at which the object is moved, meaning that the faster the object is moved, the more work is done.

## 5. Can the work done in lifting and lowering an object be negative?

Yes, the work done in lifting and lowering an object can be negative. This occurs when the object is being lowered and the force of gravity is acting in the same direction as the movement, resulting in a negative value for work done.

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