# Potential energy and lifting an object vertically

• B
• Chenkel

#### Chenkel

So the potential energy of an object in a gravitational field is pe=hmg where h is the height of the object in the gravitational field in meters m is the mass in kilograms and g is the acceleration in meters per second per second

I read on an answer to a question that the force to lift an object has to be greater than the weight of the object for there to be any vertical movement

My question is that to apply a force vertically equal to the weight of the object first you need to apply a force to cancel out the weight so the object stands still, then you need to apply the force again to have a vertical force equal to the weight of the object.

Doesn't this mean if we apply a force equal to twice the weight of the object over 10 meters the energy applied will be equal to 10*2*weight but the potential energy will be pe=mgh so pe=10*weight. My question is why is the energy applied not equal to the potential energy?

Any help will be greatly appreciated, thank you!

Doesn't this mean if we apply a force equal to twice the weight of the object over 10 meters the energy applied will be equal to 10*2*weight but the potential energy will be pe=mgh so pe=10*weight. My question is why is the energy applied not equal to the potential energy?

Any help will be greatly appreciated, thank you!
What happens to an object subject to an unbalanced force?

• Chenkel
What happens to an object subject to an unbalanced force?
To have unbalanced forces means that the force applied in one direction is greater than the force applied in the opposite direction. When unbalanced forces are acting on an object, there is a change in speed and/or direction.

So there seems to be unbalanced forces, I see a total force of 2*weight being applied through a distance of 10 meters, but the potential energy only changed by 10*weight, I suppose after going 10 meters the object may still be in vertical motion so there may be a transformation into some kinetic energy from pushing it upward and it may travel higher than 10 meters. The exact mechanics of what is going on is not very clear to me. Thanks for any light you might be able to shed on this

There is no reason to expect that the work done by an external force to be equal to the change in potential energy. The change in (gravitational) potential energy is related only to the work done by gravity. No matter what work is done by the lifting force, the work done by gravity is -mgh and the potential energy invcrease by mgh. If there is work done by other forces, like your lifting force, if the net (total) work is not zero there will be a change in the kinetic energy of the object. But this does not affect the change in potential energy which is mgh as long as the weight is mg and the displacement is h. It does not matter how the body made it through this displacement h. Maybe it moved very fast, under the effect of a very large lifting force or very slowly under a force barely larger than the weight. Or even under a zero net force, once the object gets some initial upward velocity. The PE increase by mgh in all cases.

• russ_watters and Chenkel
To have unbalanced forces means that the force applied in one direction is greater than the force applied in the opposite direction. When unbalanced forces are acting on an object, there is a change in speed and/or direction.

So there seems to be unbalanced forces, I see a total force of 2*weight being applied through a distance of 10 meters, but the potential energy only changed by 10*weight, I suppose after going 10 meters the object may still be in vertical motion so there may be a transformation into some kinetic energy from pushing it upward and it may travel higher than 10 meters. The exact mechanics of what is going on is not very clear to me. Thanks for any light you might be able to shed on this
The object accelerates and, when it reaches the height of ##10m##, has gained Kinetic Energy in addition to its Gravitational Potential Energy.

• Delta2 and Chenkel
There is no reason to expect that the work done by an external force to be equal to the change in potential energy. The change in (gravitational) potential energy is related only to the work done by gravity. No matter what work is done by the lifting force, the work done by gravity is -mgh and the potential energy invcrease by mgh. If there is work done by other forces, like your lifting force, if the net (total) work is not zero there will be a change in the kinetic energy of the object. But this does not affect the change in potential energy which is mgh as long as the weight is mg and the displacement is h. It does not matter how the body made it through this displacement h. Maybe it moved very fast, under the effect of a very large lifting force or very slowly under a force barely larger than the weight. Or even under a zero net force, once the object gets some initial upward velocity. The PE increase by mgh in all cases.
Thank you to both of you for the replies. To your answer, does this mean that all changes of potential energy are a result of transfer of kinetic energy because height changes with vertical velocity?

So if I move the object very fast vertically this force that I apply will change the velocity through a change in acceleration and accordingly change the height, it might even go higher once I release the object, at the point the object is at it's heighest all kinetic energy will have been converted into potential energy. Am I seeing things correctly?

Thank you to both of you for the replies. To your answer, does this mean that all changes of potential energy are a result of transfer of kinetic energy because height changes with vertical velocity?

So if I move the object very fast vertically this force that I apply will change the velocity through a change in acceleration and accordingly change the height, it might even go higher once I release the object, at the point the object is at it's heighest all kinetic energy will have been converted into potential energy. Am I seeing things correctly?
I haven't seen my reply on a different device after 25 minutes. Hopefully things are updating correctly

I am not sure what you mean by "height changes with vertical velocity". Of course, moving up means there is some vertical velocity. But this velocity may be constant during the upward motion so no, not all changes in potential energy are the result of transfer of kinetic energy. But it is also a matter of what you mean by "result". If a change in PE is accompanied by a change in KE, will you say that the one is the result of the other? Why not the change in KE is the result of change in PE? You may even have both KE and PE decerasing at the same time, like a ball kicked downwards on an inclined plane with high friction. I suppose I mean: don't try to make the rules more general than they are supposed to be. Use the work energy theorem and the definition of potential energy as they are.

Your second paragraph though looks pretty good.

• Chenkel
I am not sure what you mean by "height changes with vertical velocity". Of course, moving up means there is some vertical velocity. But this velocity may be constant during the upward motion so no, not all changes in potential energy are the result of transfer of kinetic energy. But it is also a matter of what you mean by "result". If a change in PE is accompanied by a change in KE, will you say that the one is the result of the other? Why not the change in KE is the result of change in PE? You may even have both KE and PE decerasing at the same time, like a ball kicked downwards on an inclined plane with high friction. I suppose I mean: don't try to make the rules more general than they are supposed to be. Use the work energy theorem and the definition of potential energy as they are.

Your second paragraph though looks pretty good.
Thank you for the reply nasu. I'm wondering why I'm not able to see all replies across devices, I've shared this thread with some friends and they aren't seeing all the posts. Is it possible that because I made my account today that the posts need to be reviewed by admins? I like this forum and would like to investigate what's not working, thank you!

I don't know about the problem with the posts. You don't need to be a signed-in memeber to see the posts. You may ask one of the moderators or the forum administrator.

• Chenkel
Think of throwing a ball. You throw it up. It reaches a peak height, then begins to fall again. What is happening with the potential and kinetic energies during the flight of the ball?

• Chenkel
I will check it out with the administrators per your advice. To your answer, how could PE change without a change in KE? Any change in height seems to change PE, and any change in height results from a change in velocity which means KE has changed. Also with conservation of energy PE is a function of KE because PE = TE - KE. Let me know what you think. Thank you.

My question is that to apply a force vertically equal to the weight of the object first you need to apply a force to cancel out the weight so the object stands still, then you need to apply the force again to have a vertical force equal to the weight of the object.
Not quite. Put a small object in the palm of your hand and lift it up at constant speed. Since the speed is constant, the kinetic energy does not change. This means that the force exerted by your hand is equal in magnitude and opposite in direction to the force of gravity. Look at it this way

Both gravity and your hand do work on the object. The two works add to zero, i.e. ##W_{hand}+W_{grav}=0## and the kinetic energy doesn't change. Since the work done by gravity is the negative of the change in potential energy (##W_{grav.}=-\Delta U_{grav.}## and the sum of the works is zero as we have seen, $$0=W_{hand}+W_{grav.}=W_{hand}-\Delta U_{grav.} \implies W_{hand}=\Delta U_{grav.}$$In other words, the work done by the hand increases the potential energy while there is no change in kinetic energy. If the hand exert an upwards force in addition to the one it's already exerting, the object will accelerate and its speed and kinetic energy will increase.

• phinds, PeroK, Chenkel and 1 other person
Not quite. Put a small object in the palm of your hand and lift it up at constant speed. Since the speed is constant, the kinetic energy does not change. This means that the force exerted by your hand is equal in magnitude and opposite in direction to the force of gravity. Look at it this way

Both gravity and your hand do work on the object. The two works add to zero, i.e. ##W_{hand}+W_{grav}=0## and the kinetic energy doesn't change. Since the work done by gravity is the negative of the change in potential energy (##W_{grav.}=-\Delta U_{grav.}## and the sum of the works is zero as we have seen, $$0=W_{hand}+W_{grav.}=W_{hand}-\Delta U_{grav.} \implies W_{hand}=\Delta U_{grav.}$$In other words, the work done by the hand increases the potential energy while there is no change in kinetic energy. If the hand exert an upwards force in addition to the one it's already exerting, the object will accelerate and its speed and kinetic energy will increase.
So there are two forces on the object, the hand, and the gravitational force, and they cancel each other out because KE is constant.

I am trying to understand negative work with some difficulty, how does gravity do work exactly? I see that you have to work against gravity so maybe that's why the work of gravity is considered negative, it's a little bit of a confusing concept to me, so please let me know if there's a more clear way of understanding it, thanks!

I will check it out with the administrators per your advice. To your answer, how could PE change without a change in KE? Any change in height seems to change PE, and any change in height results from a change in velocity which means KE has changed. Also with conservation of energy PE is a function of KE because PE = TE - KE. Let me know what you think. Thank you.
Again, you try to make some special cases into general rules. A change in height does not necesarily imply a change in velocity. Just some velocity, which may be constant.
And the total mechanical energy is conserved just for some special cases, isolated systems. In your case, with a lifting force moving the object upwards, this is not the case. The total mechanical energy is not conserved.

• PeroK and Chenkel
So there are two forces on the object, the hand, and the gravitational force, and they cancel each other out because KE is constant.
Correct.
I am trying to understand negative work with some difficulty, how does gravity do work exactly? I see that you have to work against gravity so maybe that's why the work of gravity is considered negative, it's a little bit of a confusing concept to me, so please let me know if there's a more clear way of understanding it, thanks!
To understand negative work, you have to look at the definition of the work done by a force ##F## on an object when the object is displaced by an amount ##d##. In the simplest case of a constant force, the work done by a force on an object is the product of three quantities, the magnitude of the force ##F##, the magnitude of the displacement ##d## and the cosine of the angle ##\theta## between the displacement vector and the force vector, $$W=F~d~\cos\theta.$$Since the two magnitudes are always positive, the work will have the same sign as the cosine of the angle. You should know from trig that the cosine is positive when ##0 \leq \theta < 90^o## and negative when ##90^o < \theta \leq 180^o##.

You should also understand that when a force does positive work on an object, the kinetic energy increases and when it does negative work the kinetic energy decreases. For example, when you toss a mass straight up in the air, the force of gravity (down) is opposite to the displacement (up). The cosine is ##\cos(180^o)=-1.## Gravity does negative and the kinetic energy decreases until it becomes zero. Then the object reverses direction, gravity and displacement are in the same direction, the cosine is +1 and the kinetic energy increases.

• Lnewqban and Chenkel
Correct.

To understand negative work, you have to look at the definition of the work done by a force ##F## on an object when the object is displaced by an amount ##d##. In the simplest case of a constant force, the work done by a force on an object is the product of three quantities, the magnitude of the force ##F##, the magnitude of the displacement ##d## and the cosine of the angle ##\theta## between the displacement vector and the force vector, $$W=F~d~\cos\theta.$$Since the two magnitudes are always positive, the work will have the same sign as the cosine of the angle. You should know from trig that the cosine is positive when ##0 \leq \theta < 90^o## and negative when ##90^o < \theta \leq 180^o##.

You should also understand that when a force does positive work on an object, the kinetic energy increases and when it does negative work the kinetic energy decreases. For example, when you toss a mass straight up in the air, the force of gravity (down) is opposite to the displacement (up). The cosine is ##\cos(180^o)=-1.## Gravity does negative and the kinetic energy decreases until it becomes zero. Then the object reverses direction, gravity and displacement are in the same direction, the cosine is +1 and the kinetic energy increases.
I'm a little confused about displacement, for example when a force of 2 is applied to an object over a distance of 10 meters you expect the work done to be 20 joules. Is the distance it travels displacement? In the example you gave of an object being thrown in the air the displacement seems to be in direction of velocity. Is displacement something like the net change in velocity over time? That seems very close to what you would call the position of the object, let me know what you think, thank you!

displacement
Displacement describes the net change in position (during a particular interval) It is typically the ending (vector) position minus the beginning (vector) position. It is therefore equal to {average (vector) velocity} X time

• Chenkel
I'm a little confused about displacement, for example when a force of 2 is applied to an object over a distance of 10 meters you expect the work done to be 20 joules. Is the distance it travels displacement? In the example you gave of an object being thrown in the air the displacement seems to be in direction of velocity. Is displacement something like the net change in velocity over time? That seems very close to what you would call the position of the object, let me know what you think, thank you!
Displacement is not the same as distance moved. If you through a ball up and it goes 5 metres up and then falls 5 metres back down, then it has traveled 10 metres in total. But its displacement is zero when it is back where it started.

Note also that gravity has done zero work on the ball in this case, as it ends with the same KE as it started with.

• Chenkel and hutchphd
Your last sentence is a little ambigous (the "as it" part in "as it ends with the same KE"). The gravity does zero work because it ends at the same height. It would be zero even if it ended with a different KE. It can loose energy to air drag or it may have a partially elastic collision with the ceiling. It does not matter what the final KE is. Or think about a bouncing ball dropped on floor. Each time it reaches the floor has the same PE but the KE decreases after each bounce.
Of course, for a simple case like the one you discuss, it happens that the KE is the same. But this has no influence on the work done by gravity or on the value of PE.

• • Chenkel, PeroK and Delta2
So work is a measure of how much energy a force has contributed in the direction of displacement, if a point mass moves in circular motion from (1, 0) to (0, 1) by angle equal to PI/2 where gravity is pushing down on the y axis, the displacement is sqrt(2) so the necessary work the force would have to do to move it in the quarter circle to it's final position is
W= F*sqrt (2)*cos(theta) if F is colinear with displacement then theta will be 0, so W=F*sqrt(2)

The work done by gravity is W = g*m*sqrt(2)*cos(3pi/4)

If the work done by gravity is canceled out by the work done by the force collinear to displacement then

0=F*sqrt(2) + g*m*sqrt (2)*cos(3pi/4)

Solving for F, a force of magnitude
F = -g*m*cos(3pi/4) is applied to the point mass as it moves in a circular way from (1, 0) to (0, 1) to reach equilibrium at (0, 1)

I may have made a mistake with my math as these concepts are pretty new to me, let me know what you guys think, thanks!

So work is a measure of how much energy a force has contributed in the direction of displacement, if a point mass moves in circular motion from (1, 0) to (0, 1) by angle equal to PI/2 where gravity is pushing down on the y axis, the displacement is sqrt(2) so the necessary work the force would have to do to move it in the quarter circle to it's final position is
W= F*sqrt (2)*cos(theta) if F is colinear with displacement then theta will be 0, so W=F*sqrt(2)

The work done by gravity is W = g*m*sqrt(2)*cos(3pi/4)

If the work done by gravity is canceled out by the work done by the force collinear to displacement then

0=F*sqrt(2) + g*m*sqrt (2)*cos(3pi/4)

Solving for F, a force of magnitude
F = -g*m*cos(3pi/4) is applied to the point mass as it moves in a circular way from (1, 0) to (0, 1) to reach equilibrium at (0, 1)

I may have made a mistake with my math as these concepts are pretty new to me, let me know what you guys think, thanks!

I am looking at this a little later and I'm not sure if this math makes sense, maybe if the point mass is moving in a linear way from (1, 0) to (0, 1) then the math might make a little sense, I think I should go back to the drawing board until I have a better idea of what I'm doing :)

If we have a constant force ##\vec F## and a displacement ##\vec d## then the work is always given by the dot product $$W= \vec F \cdot \vec d$$This can be thought of in two equivalent ways
1. The component of ##\vec F## along ##\vec d##
2. The component of ##\vec d## along ##\vec F##
Either way the value is $$W=Fd \cos\theta$$

• Delta2 and Chenkel
That's of course only true when the force is constant along the trajectory of the particle and if the force is conservative, i.e., derivable from a potential. The general formula for the work done within a time interval ##(t_1,t_2)## is
$$\Delta W=\int_{t_1}^{t_2} \mathrm{d} t \vec{v}(t) \cdot \vec{F}[\vec{x}(t)],$$
where ##\vec{x}(t)## is the trajectory of the particle under the influence of the force ##\vec{F}(\vec{x})## and ##\vec{v}=\dot{\vec{x}}## the velocity of the particle.

If ##\vec{F}## is conservative then there is a scalar potential ##V## such that
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
and then the work is independent of the path chosen to calculate it, i.e., you don't need to know the trajectory of the particle, and then you get
$$\Delta W=-[V(\vec{x}_2)-V(\vec{x}_1)],$$
where ##\vec{x}_1## and ##\vec{x}_2## are the (fixed) initial and final position of the particle.

For a constant force you have
$$\vec{V}=-\vec{F} \cdot \vec{x},$$
and then indeed
$$\Delta W=\vec{F} \cdot (\vec{x}_2-\vec{x}_1) = \vec{F} \cdot \vec{d}.$$

• Chenkel and hutchphd
That's of course only true when the force is constant along the trajectory of the particle and if the force is conservative, i.e., derivable from a potential.
I am slightly confused. Are you saying the first part is not sufficient?

in post #24 when @vanhees71 claims
and then the work is independent of the path chosen to calculate it, i.e., you don't need to know the trajectory of the particle,
he makes implicit use of the gradient theorem of vector calculus.

• vanhees71
I am slightly confused. Are you saying the first part is not sufficient?
Since a constant force is derivable from a potential, it's sufficient.

• hutchphd and Delta2
That's of course only true when the force is constant along the trajectory of the particle and if the force is conservative, i.e., derivable from a potential. The general formula for the work done within a time interval ##(t_1,t_2)## is
$$\Delta W=\int_{t_1}^{t_2} \mathrm{d} t \vec{v}(t) \cdot \vec{F}[\vec{x}(t)],$$
where ##\vec{x}(t)## is the trajectory of the particle under the influence of the force ##\vec{F}(\vec{x})## and ##\vec{v}=\dot{\vec{x}}## the velocity of the particle.

If ##\vec{F}## is conservative then there is a scalar potential ##V## such that
$$\vec{F}(\vec{x})=-\vec{\nabla} V(\vec{x}),$$
and then the work is independent of the path chosen to calculate it, i.e., you don't need to know the trajectory of the particle, and then you get
$$\Delta W=-[V(\vec{x}_2)-V(\vec{x}_1)],$$
where ##\vec{x}_1## and ##\vec{x}_2## are the (fixed) initial and final position of the particle.

For a constant force you have
$$\vec{V}=-\vec{F} \cdot \vec{x},$$
and then indeed
$$\Delta W=\vec{F} \cdot (\vec{x}_2-\vec{x}_1) = \vec{F} \cdot \vec{d}.$$
Vanhees, I'm a little rusty on my vector calc, what does del with a line over it mean, and why is that operation on V equal to F, what is V?

It's the nabla operator. In Cartesian coordinates it's
$$\vec{\nabla}=\begin{pmatrix}\partial/\partial x\\ \partial/\partial y \\ \partial/\partial z \end{pmatrix}.$$

• Chenkel
how does gravity do work exactly? I see that you have to work against gravity so maybe that's why the work of gravity is considered negative,
I frequently read about confusion where 'doing work' is discussed. I think there's a bit of innate anthropomorphism involved in this. Someone (or some agency) 'has' to be doing something so that brings in the 'done by' and 'done to' idea.
There is no need for confusion as long as force and displacement are described consistently (magnitude and direction). When there are problems in static mechanics, people tell you to draw a free body diagram and that will get over any clash with intuition. The work idea can be treated in just the same way and will give consistent results - e.g. a mass on a spring will fall down due to gravity. No problem to describe that happening but there's a temptation to say gravity is 'doing the work' on the way down and the spring is 'doing the work' on the way up. A sudden change of reference like that has got to spell danger.

• Chenkel, vanhees71 and Lnewqban