The discussion clarifies the distinction between work done by a force and work done against a force, specifically in the context of gravitational force. Work done by gravity is represented as W = -mgh, while work done against gravity, such as when lifting an object, is calculated as W = mgh. The conversation emphasizes that work done against a force is the energy required to move an object in opposition to that force, and it is crucial to differentiate this from work that contributes to kinetic energy. The relationship between the two types of work is that they are equal in magnitude but opposite in sign.
PREREQUISITESStudents of physics, educators teaching mechanics, and anyone interested in understanding the principles of work and energy in physical systems.
For example, we can easily calculate the work done by the force of gravity in moving an object up or down an inclined plane by using the expressions W=FSCostheta or W=(-)mgh but how would we calculate the work done against the force or gravity? And what exactly does 'work done against the force of gravity' (or friction or any other force for that matter) mean?Doc Al said:Sounds like a bit of semantics. Can you give some examples of exactly what you mean?
Also, if work IS being done against a force (gravity, friction, etc), what exactly is doing the work?Doc Al said:Sounds like a bit of semantics. Can you give some examples of exactly what you mean?
The terminology is a bit loose, but prevalent. Here's an example: You lift a box (at constant speed) a distance h. There are two forces acting on the box: gravity and the force exerted by you. The work done by gravity (as you well know) is -mgh; the work done by you, which you could think of as work done 'against gravity' is mgh.D. Wani said:For example, we can easily calculate the work done by the force of gravity in moving an object up or down an inclined plane by using the expressions W=FSCostheta or W=(-)mgh but how would we calculate the work done against the force or gravity? And what exactly does 'work done against the force of gravity' (or friction or any other force for that matter) mean?
How is work done by the person equal to mgh?Doc Al said:The terminology is a bit loose, but prevalent. Here's an example: You lift a box (at constant speed) a distance h. There are two forces acting on the box: gravity and the force exerted by you. The work done by gravity (as you well know) is -mgh; the work done by you, which you could think of as work done 'against gravity' is mgh.
But when work is involved, there is always something exerting a force on something else.
The force they exert = mg and the distance over which they exert it = h. Work = mgh.D. Wani said:How is work done by the person equal to mgh?
Okay, so basically, the work done on a body against a force is the work required to move the body against the force? But just the work that is used in moving the body against the force, not the work that accelerates the body or causes its kinetic energy to increase. Did I get it right?Doc Al said:The force they exert = mg and the distance over which they exert it = h. Work = mgh.
If they exert a force greater than mg, then the work they do 'against gravity' would be the same, but they will be doing additional work to increase the kinetic energy of the box.
So when we say work is being done against a force, we are simply saying that work is being done on a force in the presence of another force which is directed in the opposite direction. Is this correct?IgorIGP said:The work done by the force F1 is done against a force F2 when the forces F1 and F2 are directed in opposite sides one to other.
Thank youQuantumQuest said:Simply put and according to Thermodynamics: if you supply energy through work, then work is taken as positive and when work is done by the system, it is taken as negative. If you have various forces on a system, the whole thing boils down to what force you're talking about. Producing or consuming? Then work has the appropriate sign.
Also, how is the force the person exerts=mg.Doc Al said:The force they exert = mg and the distance over which they exert it = h. Work = mgh.
If they exert a force greater than mg, then the work they do 'against gravity' would be the same, but they will be doing additional work to increase the kinetic energy of the box.
As long as the person just holds the object, yes.D. Wani said:Also, how is the force the person exerts=mg.
In order to move the object, use a force greater then mg. When you are satisfied with its position, use a force slightly less than mg until the object is at rest.D. Wani said:So technically, the body shouldn't move?
Zero net force doesn't only mean the body shouldn't move. It means, from Newton's first law, that the body maintains it's state of motion. In this case, since both the forces cancel each other, the person moves with a constant velocity. If the person were at rest, a slightly larger force than gravity would be needed to first set him into motion. That force can later be reduced slightly to make it equal to mg, when the person attains required velocity v. Now, since the net force is 0, the person will move with constant velocity v.D. Wani said:So technically, the body shouldn't move?
When we are calculating work done, we have to calculate the work done in actually moving the body. That means the force HAS to be greater than mg, but when we are calculating the work done against gravity, we take (mg). But if the person is only applying a force of (mg) the body won't move because the gravitational force is also (mg) and they will cancel each other out.Svein said:In order to move the object, use a force greater then mg. When you are satisfied with its position, use a force slightly less than mg until the object is at rest.
Okay, so when the net force becomes zero, it keeps moving.cnh1995 said:Now, since the net force is 0, the person will move with constant velocity v.
Yes. But if the initial force is infinitesimally greater than mg and is applied for an instant only, just to break the inertia and set the body into motion, it can be considered 0. Of course, the velocity attained is also very small(→0).D. Wani said:But then wouldn't we have to keep two forces in mind while calculating the work done?
So, when we consider the force applied by the person to be zero, and we calculate work done, we are just calculating the work done in moving the body against gravity, not in increasing its kinetic energy or in accelerating it? Work done in moving a body against gravity and work done in increasing the k.e. or accelerating it are different. Is this correct?cnh1995 said:Yes. But if the initial force is infinitesimally greater than mg and is applied for an instant only, just to break the inertia and set the body into motion, it can be considered 0. Of course, the velocity attained is also very small.
Yes. It's mg(h2-h1). Force applied by the person is not 0,it's mg. Net force on the body is 0. Initial kinetic energy is also 0 as v→0. The work done will be for moving the body against the gravity and the person will attain a potential energy of mgh. If the initial velocity is u and final velocity is v and no force is applied on the body except the gravity, lost KE of the body=gained PE i.e.D. Wani said:So, when we consider the force applied by the person to be zero, and we calculate work done, we are just calculating the work done in moving the body against gravity,
No. Absolutely. It does not metter were the body is moving. Sadnifficant is only:D. Wani said:the work done on a body against a force is the work required to move the body against the force?
And it means that the F1 works aganist to F2 and F2 works aganist to F1. Have you ever seen an arm-wrestling fight?IgorIGP said:when the forces F1 and F2 are directed in opposite sides one to other
We are concluding all of this assuming friction and air resistance are absent, right?cnh1995 said:When the body is set into motion and the net force on the body becomes 0, it will move up with constant velocity v. The actual energy at a height h will be E=mgh+mv2/2. But since v→0 as the initial force was infinitesimally greater than mg and was applied just for an instant,
E=mgh+lim(v→0) mv2/2.
∴E=mgh.
I got a little confused?IgorIGP said:No. Absolutely. It does not metter were the body is moving. Sadnifficant is only:
And it means that the F1 works aganist to F2 and F2 works aganist to F1. Have you ever seen an arm-wrestling fight?
Yes. Everything is assumed to be ideal.D. Wani said:We are concluding all of this assuming friction and air resistance are absent, right?
You do. Do not think about friction or resistance or reactions yet. If the force makes work and exists anoter force that is directed in opposit to it, that means that both work aganist one to each other. That is much more easily than you think now. Just relax and imagine the arm-wrestling.D. Wani said:I got a little confused?
When the net force is zero, the velocity is constant, not necessarily zero. If you lift the box at constant speed you'll need to exert a force equal to its weight.D. Wani said:Also, how is the force the person exerts=mg.
Technically, the gravitational force=mg. If the force the person is applying is equal to mg, then the gravitational force and the person's force would cancel out leaving a net force of zero. So technically, the body shouldn't move?
But, as cnh1995 said, for the body to be set into motion, the initial force has to be in infinitesimally greater than mg, just at that instant.. After that, to maintain constant speed, we need to apply a force equal to (mg). Right?Doc Al said:When the net force is zero, the velocity is constant, not necessarily zero. If you lift the box at constant speed you'll need to exert a force equal to its weight.
Though work is only done when there is displacement of the bodyIgorIGP said:You do. Do not think about friction or resistance or reactions yet. If the force makes work and exists anoter force that is directed in opposit to it, that means that both work aganist one to each other. That is much more easily than you think now. Just relax and imagine the arm-wrestling.