What is a Newton? Is it correctly understood?

  • #1

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Hi,
If one Newton is defined as the amount of force required to give a 1kg mass an acceleration of 1m/s/s is it then correct to say that "10 newton is the amount of force required to give a 10kg mass an acceleration of 1m/s/s" and would it be equally as right to say "10 newton is the amount of force required to give a 1kg mass an acceleration of 10m/s/s"

Help is appreciated :)
Christian
 

Answers and Replies

  • #2
Nugatory
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yes.
 
  • #3
That's interesting because work=opposing force • distance moved = 1kg•m/s/s • m = 1kg•m^2/s/s= joules

So does this mean that one joule is the same as moving an object of mass 1kg one meter against an opposing force which makes the 1kg mass accelerate 1m/s/s in the opposite direction of which the mass was moved?
Or does it mean
"moving an object of ANY MASS one meter against an opposing force which makes the 1kg mass accelerate 1m/s/s in the opposite direction of which the mass was moved?"

Thanks again :)
 
  • #4
Nugatory
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So does this mean that one joule is the same as moving an object of mass 1kg one meter against an opposing force which makes the 1kg mass accelerate 1m/s/s in the opposite direction of which the mass was moved?
Or does it mean
"moving an object of ANY MASS one meter against an opposing force which makes the 1kg mass accelerate 1m/s/s in the opposite direction of which the mass was moved?"
The latter is much closer to correct, but you really should not be thinking in terms of an opposing force at all. Just say that you are applying a force of one Newton to the object as it moves. By Newton's third law, you won't be able to do this unless there is an equal and opposite force on you (which is, I think, what you're calling the "opposing force") but it's the force that you're applying that is doing the work.

It is sometimes hard to visualize exactly what is meant by pushing with a constant force of one Newton, but here's one way of thinking about it: Place an uncompressed coil spring with a spring constant of 1N/cm between your hand and the object, then press on the spring to compress it by one cm. As long as you move your hand around in a way that maintains that one cm compression, you are applying one Newton of force, and when you've moved your hand one meter you'll have done one Joule of work on the object. If the object is very heavy this may take a very long time and it won't be moving very quickly when you're done; if it's light enough you'll have to move your hand so quickly that you're almost throwing the object as it accelerates away from you.
 
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  • #5
A.T.
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Or does it mean,,,

Why don't you just plug numbers into the equations to see what they mean? They are as simple as it gets,
much shorter than convoluted questions which can be easily misinterpreted.
 
  • #6
It's just because in thermodynamics work means opposing force • distance moved, that's why I use "opposing force" since I'm studying thermodynamics ( forgot to mention that) so on terms of thermodynamics is the latter statement for a joule correct?


A.T
I don't underatand? The expression of a joule has the number ONE in front of all the units so there is already a number in frot of it, I'm just trying to interpret it so I can really understand how a joule is meant to be understood and how it would look visually.
 
  • #7
russ_watters
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Where exactly did you see the term "opposing force" used in that way? It looks wrong.
 
  • #8
ehild
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Speaking about work you must specify what does the work and on what.
The gas does work on the piston which is equal the force it exerts on the piston multiplied by the displacement of the piston.

An external force F does work on the piston which is equal to the force times displacement again. If the piston has no mass or the process is quasistatic, the piston does not take energy and the whole work is done on the gas.

As Nugatory suggested, you should forget that opposing force. Work done on an object is equal to the force exerted on the object multiplied by the displacement of the object.
 
  • #9
Speaking about work you must specify what does the work and on what.
The gas does work on the piston which is equal the force it exerts on the piston multiplied by the displacement of the piston.

An external force F does work on the piston which is equal to the force times displacement again. If the piston has no mass or the process is quasistatic, the piston does not take energy and the whole work is done on the gas.

As Nugatory suggested, you should forget that opposing force. Work done on an object is equal to the force exerted on the object multiplied by the displacement of the object.
According to the book by Peter Atkins "chemical principles - the quest for insight" the force that is exerted on an object in order to move it some distance must e counteracted by some force or else you could move it an infinite distance.

But It must be true that if one Newton is the amount of force required to give a 1kg mass anacceleration of 1m/s/s then if we multiply that with distance to get work, then one joule of work must be the force required to accelerate a mass of 1kg through one meter distance?

If this is not true the what is te true definition of a jule? There must be a correct and detailed definition of the joule which has some type of interpretable meaning?
 
  • #10
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According to the book by Peter Atkins "chemical principles - the quest for insight" the force that is exerted on an object in order to move it some distance must e counteracted by some force or else you could move it an infinite distance.

But It must be true that if one Newton is the amount of force required to give a 1kg mass anacceleration of 1m/s/s then if we multiply that with distance to get work, then one joule of work must be the force required to accelerate a mass of 1kg through one meter distance?

If this is not true the what is te true definition of a jule? There must be a correct and detailed definition of the joule which has some type of interpretable meaning?
No, definitely not. The joule cannot be a force. Any kind of force. They measure different things.

And the first paragraph here is really bad. And not true. You cannot move anything an infinite distance even if there is no "counetracting" force. And you can have forces without counteracting forces. And anyway it has nothing to do with definition of work.
 
  • #11
ehild
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I joule is the work done by 1 N force applied at an object when it displaces by 1 m in the direction of the force.
In general the work of force F during the displacement D is W=FD cos(θ) where θ is the angle between the force and displacement.
It does not matter if the object accelerates or not. If there are other forces acting on the same object they also do work. It is possible that the net work is zero. In that case, the object moves with constant velocity.
Imagine you let a ball of 1kg fall by 1 m. The force of gravity is 9,8 N. It does 9.8 J work.
Now you hold the same ball and move downward very slowly by 1 m. The ball moves with constant velocity, which means your force balances gravity. The force of gravity is the same as before, and so is the work of gravity.
Your force is 9.8 N upward. It is opposite to the direction of the displacement, so you do negative work: -9.8 J.

The net force applied on the ball is zero, and the net work done on it is also zero.
 
  • #12
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Ehild: "Imagine you let a ball of 1kg fall by 1 m. The force of gravity is 9,8 N. It does 9.8 J work.
Now you hold the same ball and move downward very slowly by 1 m. The ball moves with constant velocity, which means your force balances gravity. The force of gravity is the same as before, and so is the work of gravity.
Your force is 9.8 N upward. It is opposite to the direction of the displacement, so you do negative work: -9.8 J.
The net force applied on the ball is zero, and the net work done on it is also zero."

The ball is losing gravitational potential energy as it drops. As a result, the ball is doing work on your hand as it moves downward.
Your description of a person doing "negative work" in this scenario is confusing. Does this mean that the person is gaining energy of some form? What form?

As well, this example adds the all pervasive force of gravity into the mix, so we are now faced with dealing with Newton's Third Law (action and reaction forces acting on different objects).
 
  • #13
russ_watters
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According to the book by Peter Atkins "chemical principles - the quest for insight" the force that is exerted on an object in order to move it some distance must e counteracted by some force or else you could move it an infinite distance.
That's a mess. Please provide the actual quote. Two important clarifications:
1. Any single force applied to an object for any amount of time will cause the object to move an infinite distance. This is a consequence of Newton's laws of motion: an "unbalanced force" causes an acceleration (Newton's 2nd law) and an object in motion will continue in motion forever unless a force is applied to stop it (Newton's 1st law). http://en.wikipedia.org/wiki/Newton's_laws_of_motion

2. But so what? That has nothing to do with work. Work is force times the distance moved while the force is being applied.
But It must be true that if one Newton is the amount of force required to give a 1kg mass anacceleration of 1m/s/s then if we multiply that with distance to get work, then one joule of work must be the force required to accelerate a mass of 1kg through one meter distance? [emphasis added]
Sorry, but that is gibberish. Force is not work. They are two completely differen things.
If this is not true the what is te true definition of a jule? There must be a correct and detailed definition of the joule which has some type of interpretable meaning?
A joule is a unit of work equal to 1 N-m or, breaking down the Newton, 1 kg-m^2/s^2
 
  • #14
russ_watters
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The ball is losing gravitational potential energy as it drops. As a result, the ball is doing work on your hand as it moves downward.
Your description of a person doing "negative work" in this scenario is confusing. Does this mean that the person is gaining energy of some form? What form?
If people were springs, they could store energy in this fashion, but they aren't. They are really really inefficient machines that require input energy even to absorb work.
 
  • #15
Andrew Mason
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That's a mess. Please provide the actual quote. Two important clarifications:

Sorry, but that is gibberish. Force is not work. They are two completely different things.
To be fair, it seems to me that the OP understands that work is a force applied through a distance:

Christian0710 said:
But It must be true that if one Newton is the amount of force required to give a 1kg mass an acceleration of 1m/s/s then if we multiply that with distance to get work, ...
Christian0710 said:
then one joule of work must be the force required to accelerate a mass of 1kg through one meter distance?
As you point out this is a bit sloppy. One Joule of work is done when a force required to accelerate a mass of 1kg at a rate of 1 m/sec2 is applied through one meter of distance.

AM
 
  • #16
Andrew Mason
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ehild said:
As Nugatory suggested, you should forget that opposing force. Work done on an object is equal to the force exerted on the object multiplied by the displacement of the object.
In thermodynamics the opposing force is important. For example, the work done by a gas during an irreversible expansion is determined by the external pressure (ie. the pressure opposing expansion of the gas) x change in volume. The work done by the surroundings in compressing a gas is determined not by the pressure of the surroundings but by the pressure of the gas (i.e. the pressure opposing compression) x change in volume.

AM
 
  • #17
ehild
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@AM: you are not right. The work done on the gas by the surroundings is the external force times the displacement of the point of application. Only in case of a quasistic process when the external and internal forces are balanced, when the external pressure and the pressure of the gas are equal, is the work on the gas equal to the pressure of the gas multiplied by the negative change of volume, W = -PΔV.
You can compress a gas with a force much bigger than its initial pressure. In that case, the work is not equal to -PdV. As it is a non-equilibrium process, the pressure of the gas is not defined during the compression.


ehild
 
  • #18
Andrew Mason
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@AM: you are not right. The work done on the gas by the surroundings is the external force times the displacement of the point of application. Only in case of a quasistic process when the external and internal forces are balanced, when the external pressure and the pressure of the gas are equal, is the work on the gas equal to the pressure of the gas multiplied by the negative change of volume, W = -PΔV.
You can compress a gas with a force much bigger than its initial pressure. In that case, the work is not equal to -PdV. As it is a non-equilibrium process, the pressure of the gas is not defined during the compression.
When a gas expands rapidly (i.e. irreversibly) against an external pressure the work done on the surroundings is determined by the external pressure, not the pressure of the gas. So when applying the first law: [itex]W = Q - \Delta U[/itex] where W is the work done by the gas, one uses the opposing pressure x change in volume to determine W. If a gas with internal pressure > 0 is allowed to freely expand into a vacuum, the work done is 0 because the opposing pressure is 0.

The compression of a gas can only be done by a constant external pressure if that pressure is greater than the initial pressure of the gas. The compression ends when the gas pressure and external pressures are equal. The pressure of the gas will not be constant during compression so the work done is never PgasΔV. Rather [itex]W = \int P_{gas}dV[/itex].

The pressure of the gas can be defined during a non-quasi-static compression. Think of a large volume of air in the atmosphere being compressed into a tank by very small pump. The pressure of the tank is defined during each cycle of the pump and it is that pressure that determines the work done.

AM
 
  • #19
ehild
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The pressure of the gas can be defined during a non-quasi-static compression. Think of a large volume of air in the atmosphere being compressed into a tank by very small pump. The pressure of the tank is defined during each cycle of the pump and it is that pressure that determines the work done.

AM
That is called a quasi-static process. :)
 
  • #20
russ_watters
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To be fair, it seems to me that the OP understands that work is a force applied through a distance:
I'm not sure I agree -- it almost seems to me like a deliberate attempt to twist the definition into something else.
 
  • #21
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Do we remember christian0710's original question?
Yes, your original statement about the definition of 1 newton and acceleration is correct BUT remember, the force applied refers to an unbalanced force. So there is no friction or other force opposing the motion in your original statement.
So, be careful about using just the word "force" in your thinking.
If you apply a force on a block of 1 N and while sliding, and the opposing friction is 1 N, then the unbalance force is 0 and no acceleration results (although the block continues to move at constant velocity as long as...)
Some of the examples in this thread may cause more confusion than understanding for you. For instance, if you hold a 1 kg mass in your hand without moving vertically, your hand exerts an upward force just balancing gravity on that mass. Even though no work is being done in the classical sense, your arm eventually gets tired. This is because you are actually expending energy through the natural rapid expansion/compression of the muscle fibres in your arm but this does not translate into any gain/loss of gravitational potential energy nor change in kinetic energy of the mass.

In Newton's First and Second Laws, the "force" referred to in their statements is an unbalanced force.
 
  • #22
Andrew Mason
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That is called a quasi-static process. :)
Well, not exactly. It does not take an arbitrarily long time to make the compression. I used that example (the volume of the pump is small compared to the volume of the tank so the pressure against which the pump is acting (i.e. the tank gas pressure) is practically constant during each cycle of the pump. In this case, would the work done in compressing the gas not be:
[tex]W = \sum_{i=1}^N P_i\Delta{V_i}[/tex]
where Pi is the pressure of the gas during the ith cycle and [itex]\Delta V_i[/itex] is the change in volume during the ith cycle?

AM
 
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  • #23
Andrew Mason
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Do we remember christian0710's original question?
Yes, your original statement about the definition of 1 newton and acceleration is correct BUT remember, the force applied refers to an unbalanced force. So there is no friction or other force opposing the motion in your original statement.
So, be careful about using just the word "force" in your thinking.
In an interaction between two bodies, there is a force opposing the motion of each body even in the absence of friction. The forces are unbalanced because they act on different bodies. That is to say that body A in impressing a force of one Newton on body B experiences an equal and opposite force from body B. The OP was merely saying (I think) that during the interaction between bodies A and B, the work done by body A on body B is equal to the negative of the force of body B on body A (FBonA = -mBaB where the acceleration is measured in an arbitrary inertial frame of reference) multiplied by the displacement through which that force acts. There is nothing wrong with that analysis.

AM
 
  • #24
I joule is the work done by 1 N force applied at an object when it displaces by 1 m in the direction of the force.
In general the work of force F during the displacement D is W=FD cos(θ) where θ is the angle between the force and displacement.
It does not matter if the object accelerates or not. If there are other forces acting on the same object they also do work. It is possible that the net work is zero. In that case, the object moves with constant velocity.
Imagine you let a ball of 1kg fall by 1 m. The force of gravity is 9,8 N. It does 9.8 J work.
Now you hold the same ball and move downward very slowly by 1 m. The ball moves with constant velocity, which means your force balances gravity. The force of gravity is the same as before, and so is the work of gravity.
Your force is 9.8 N upward. It is opposite to the direction of the displacement, so you do negative work: -9.8 J.

The net force applied on the ball is zero, and the net work done on it is also zero.
Thank you for the reply,
So in terms of your definition of the joule What is the difference between saying
A) "The work done by 1N force"
B) "The work done by 1N force applied at an object when it displaces 1 meter"

My confusion is this: Both quotes seem to involve ONE newton of force" Why would you want a Joule to be a newton of force applied to move some object one meter? Why not just a newton of force?
 
  • #25
A.T.
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Why would you want a Joule to be a newton of force applied to move some object one meter? Why not just a newton of force?

If a Joule was just a Newton, they would be the same thing. Why would you give two different names to the same thing?
 

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