# What is a Newton? Is it correctly understood?

1. Oct 20, 2014

### christian0710

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

2. Oct 20, 2014

### Staff: Mentor

yes.

3. Oct 20, 2014

### christian0710

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. Oct 20, 2014

### Staff: Mentor

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.

Last edited: Oct 20, 2014
5. Oct 20, 2014

### A.T.

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. Oct 20, 2014

### christian0710

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. Oct 20, 2014

### Staff: Mentor

Where exactly did you see the term "opposing force" used in that way? It looks wrong.

8. Oct 20, 2014

### ehild

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. Oct 20, 2014

### christian0710

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. Oct 20, 2014

### nasu

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. Oct 20, 2014

### ehild

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. Oct 20, 2014

### daqddyo1

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. Oct 20, 2014

### Staff: Mentor

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.
Sorry, but that is gibberish. Force is not work. They are two completely differen things.
A joule is a unit of work equal to 1 N-m or, breaking down the Newton, 1 kg-m^2/s^2

14. Oct 20, 2014

### Staff: Mentor

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. Oct 21, 2014

### Andrew Mason

To be fair, it seems to me that the OP understands that work is a force applied through a 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. Oct 21, 2014

### Andrew Mason

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. Oct 21, 2014

### ehild

@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. Oct 21, 2014

### Andrew Mason

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: $W = Q - \Delta U$ 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 $W = \int P_{gas}dV$.

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. Oct 21, 2014

### ehild

That is called a quasi-static process. :)

20. Oct 21, 2014

### Staff: Mentor

I'm not sure I agree -- it almost seems to me like a deliberate attempt to twist the definition into something else.