17
$\begingroup$

I'm not getting the definition of a joule. From the definitions I've read if I apply one newton of force to any object, now matter how heavy/ how much mass it has, over one metre in a single direction then I've done 1 joule of work.

a unit of work or energy equal to the work done by a force of one newton acting through a distance of one meter

A unit of energy equal to the work done when a force of one newton acts through a distance of one meter.

equal to the work done by a force of one newton when its point of application moves through a distance of one meter in the direction of the force:

So a force of one newton pushing on an object 1 kg heavy over one metre is 1 joule of work done, and a force of one newton pushing on an object 1000 kg heavy over one metre is 1 joule of work done? How is it the same amount of work done if I have to be pushing on that 1000 kg object a lot longer than the 1 kg object to move it over a metre?

$\endgroup$
13
  • 11
    $\begingroup$ Note that the large object will take a much longer interval of time to move the 1 meter than the small one will. Hope that helps a bit. $\endgroup$ – garyp May 11 at 10:26
  • 1
    $\begingroup$ This one in particular is probably a duplicate, but I'll refrain from using my dupe-hammer in case I'm missing something physics.stackexchange.com/q/79523/179151 $\endgroup$ – BioPhysicist May 11 at 13:08
  • 2
    $\begingroup$ “dupe-hammer” is the best thing I’ve read all day $\endgroup$ – Natru May 11 at 15:03
  • 1
    $\begingroup$ @Natru I can't take credit for it $\endgroup$ – BioPhysicist May 11 at 15:20
  • 1
    $\begingroup$ What might help you is to include power into your thoughts, which might be a more intuitive concept in this case. Since the work is the same in both cases, for the heavy and the light object, moving the heavy object with the same amount of force takes much longer and thus, the power is much lower (if linearly, it's just energy/time). $\endgroup$ – infinitezero May 13 at 19:50

12 Answers 12

24
$\begingroup$

How is it the same amount of work done if I have to be pushing on that 1000kg object a lot longer than the 1kg object to move it over a metre?

Just because it takes more time doesn't mean more energy is spent. Think of gravity. It is constantly pulling in the book on the shelf - but no energy is spent, no work is done.

Only when the book falls is work done by pulling it further down. But it doesn't matter if the book started from rest and now falls while slowly speeding up, or if the book was thrown down with high initial speed so that it reaches the ground much faster - in both cases gravity did the same amount of work, spent the same amount of energy.

The duration that the force acts over has no influence on energy being spent on the object, the work that being done.

But you might still object to this answer. It does feel tougher to push on something over a longer time. You can clearly feel more energy being spent. And you are absolutely right. But this is of another reason: this is due to our inefficient human bodies and not due to more energy being given to the pushed object.

Our human bodies spend energy just to produce a force. So pushing for a longer time does correspond to more energy being spent. But all that extra energy is not spent as work on the object, it is just spent on compressing and elongating muscle fibres, on increasing heart rate and respiration, on adrenalin flows and circulation etc. Also a lot is wasted as heat. Of this reason it is unwise to rely on the human feeling when we are talking about energy usage in science. We are too darn inefficient and so much is going on beneath our surfaces for us to be relied upon.

$\endgroup$
9
  • 3
    $\begingroup$ Suppose I framed it a slightly different way, instead of a human pushing on the object, I attach a little rocket with a thrust of one newton pushing on the object. To move it over one metre that rocket has to be burning much longer to move the 1000kg object a metre than the 1kg object. It's hard to picture that the rocket isn't doing much more work (burning more fuel also) to do that to the heavier object. $\endgroup$ – Zebrafish May 11 at 11:56
  • 7
    $\begingroup$ @Zebrafish A rocket is in the same category as the human body. It spends energy just to produce a force. It actually spends energy just by being on. If you set this rocket to push on a stationary wall, then you would see plenty of energy being spent without the wall moving anywhere - clearly, no energy has been given to the wall. But pick another force, a conservative force for instance such as gravity or such as a stretched rubber band or spring and let this force do the pulling. Then the exact same amount of energy is spent regardless of time. $\endgroup$ – Steeven May 11 at 12:03
  • 3
    $\begingroup$ I think I'm not understanding the definition. When I read the definition and it says a 'unit of work' I'm picturing the exerting of a force as being work, when clearly it isn't. Is this a discrepancy between the physics meaning of work and the ordinary English one? Because try to pick up a weight and it doesn't move I'm still creating a force on it, and doing work, at least in ordinary English. So you're saying because I don't make it move I'm not doing any work? $\endgroup$ – Zebrafish May 11 at 12:09
  • 2
    $\begingroup$ @Zebrafish " So you're saying because I don't make it move I'm not doing any work " This is correct. if you dont make it move, then no work is being done by you on the object, according to the physics definition of work $\endgroup$ – silverrahul May 11 at 12:37
  • 17
    $\begingroup$ @Zebrafish In physics terms, remember that energy is not defined as the work spent by the thing that pushes, but as work received by the thing being pushed. In other words, this definition doesn't care about what other energy you waste in order to supply energy - to do work - to the object. All the definition includes is the actual work that ends up being done on the object. Work done within muscles or energy spent just to produce the force only belong to those special cases, to those inefficient "machines" that can't produce forces without spending energy. This isn't general behaviour. $\endgroup$ – Steeven May 11 at 13:51
17
$\begingroup$

Suppose we exert a force $F$ on an object of mass $m$ for a time $t$. Then the acceleration of the object is

$$a = \frac F m$$

If the object is initially stationary then after time $t$ its velocity is

$$v = at = \frac {Ft} m$$

so the object's kinetic energy after time $t$ is

$$\frac 1 2 mv^2 = \frac 1 2 ma^2 t^2 = ma \left( \frac 1 2 at^2 \right) = F \left( \frac 1 2 at^2 \right)$$

It has also travelled a distance

$$d = \frac 1 2 a t^2$$

so

$$\text{Work done on object}= \text{gain in kinetic energy of object} = F \left( \frac 1 2 at^2 \right) = Fd$$

$\endgroup$
1
  • 3
    $\begingroup$ I think you should make it explicit that you are not starting with the definition of work given by the OP, but rather you are starting with a definition of "work is change in kinetic energy" and then showing that this is equivalent to the definition of work given by the OP. If not, then it is a pretty circular argument. $\endgroup$ – BioPhysicist May 12 at 14:26
11
$\begingroup$

There are really great answers here. However, just to cut to the heart of it all, the answer is simple.

The answer is because that is the definition of a joule and work. The definition is only concerned with a force applied over a distance. The definition doesn't concern itself with time, mass of the object the force is being applied to, etc.

Any other answer that "shows" that work is independent of mass, time, etc. is either

  1. Using circular reasoning by starting with the given definition and then arriving at the given definition, or
  2. Is implicitly starting with some other definition of work (such as it is the change in kinetic energy) and then arriving at the definition given by the OP to show independence of mass, time, etc.

Number 2 is valid, but unnecessary if we already have the definition that "solves the problem" from the beginning.

$\endgroup$
6
  • 3
    $\begingroup$ @Gert No, there is no "showing these don't matter" unless you start with a different definition and then arrive to that conclusion. What definition are you proposing I start with? I think I should start with the definition given by the OP, which at that point you're done. $\endgroup$ – BioPhysicist May 11 at 13:04
  • 1
    $\begingroup$ I think that's helpful. My conception of 'work' was wrong. For what it's worth I think the physics definition is not what my mind was thinking. $\endgroup$ – Zebrafish May 12 at 4:52
  • 1
    $\begingroup$ This answer assumes that all we are interested in is "solving the problem" rather than explaining it in a way that the OP (and other readers) will find intuitive. OP's opening sentence was "I'm not getting the definition of a joule". Answering "because that is the definition" isn't helpful here. Number 2 is only unnecessary if you have already accepted the original definition at face value, which the OP has not. If they had, then they wouldn't have needed to ask the question in the first place. $\endgroup$ – JBentley May 13 at 13:12
  • $\begingroup$ @JBentley I disagree. The issue is that the OP is not understanding that we start with the definition. They are taking their own perception of what they think work is and then asking why the definition doesn't agree with that. I do a lot of tutoring, and one of the key issues I see students have is that they do not take a definition for what it is. I think a reminder saying, "by definition this is what work is" is extremely helpful. The OP even commented above saying my answer was helpful, so I think saying this answer is unhelpful is not entirely accurate. $\endgroup$ – BioPhysicist May 13 at 14:59
  • $\begingroup$ Stackexchange exists for the benefit of all future readers, not just the OP. You've got 2 comments here along the lines of mine, and less than half the upvotes of other answers, so I think there is definitely room for improvement. I'm not saying that your approach is wrong, just that saying "Because that's the definition" with very little elaboration isn't really very helpful to someone who is confused. If we break down your answer, paragraph 1 adds nothing, paragraph 2 is your main point but scant on details, and the rest merely explains why you think the other answers are problematic. $\endgroup$ – JBentley May 13 at 16:14
3
$\begingroup$

You're conflating things.

Say we have an object of $1000\mathrm{kg}$ mass, placed on a frictionless, horizontal surface.

Now we exert a horizontal force of $1\mathrm{N}$ on it.

Newton's 2nd law now tells us that:

$$F=ma\Rightarrow a=\frac{F}{m}$$

where $a$ is the acceleration the object experiences.

Now assume that at the very moment we start applying the force of $1\mathrm{N}$ the object was still stationary ($v=0$) then its speed will evolve as:

$$v(t)=at$$

The object's displacement $x(t)$ will be:

$$x(t)=\frac12 at^2$$

The time $t$ to travel $1\mathrm{m}$ will be:

$$t=\sqrt{\frac{2}{a}}=\sqrt{\frac{2m}{F}}$$

Or for a force of $1\mathrm{N}$: $$t=\sqrt{2m}$$

In that time, $1\mathrm{J}$ will have been expended ($1\mathrm{N}\times 1\mathrm{m}$).

It's now obvious that if we use an object of lower mass $m$ we'll still expend $1\mathrm{J}$ but in a shorter amount of time.

$\endgroup$
3
$\begingroup$

I would just like to offer a perspective that doesn't focus so much directly on work/energy.

It's understandable that since the bigger mass experiences the force for a longer time, it feels like it should have more something than the smaller mass. It just turns out that something is momentum, not energy.

Taking your example of $m_1 = 1 \space kg$ and $m_2 = 1000 \space kg$, we can calculate that after the application of 1 Joule of work each mass will be traveling at $v = \sqrt{\frac{2}{m} \cdot 1J}$, so

$v_1 = 1.414 \space m/s$
$v_2 = 0.045 \space m/s$

Calculating momentum as $p = mv$ we have

$p_1 = 1.414 \space kg \space m/s$
$p_2 = 44.72 \space kg \space m/s$

So the larger mass has significantly more momentum, even though they have the same amount of energy.

Suppose instead we applied a force of 1N for 1 second to each mass. Then

$v_1 = 1\space m/s$
$v_2 = 0.001\space m/s$

Each mass now has the same amount of momentum, but if you calculate out the energy, the smaller mass has significantly more energy ($0.5J$ vs $0.0005J$)

$\endgroup$
3
$\begingroup$

Another way to intuit this is to imagine a spring. We compress the spring, set it up in a zero-friction environment, and let it push the mass:

Spring ⬇
|ssssssss|[? kg]
           ⬆ Mass

... time passes ...

         |-  1m  -|
|ssssssssssssssssss|[? kg]-->

By conservation of energy, the stored energy of the spring compression has gone into accelerating (moving) the mass. Whether it did that quickly on the 1kg mass or very very slowly on the 1000 kg mass is irrelevant - the change in the energy of the spring is the same either way, and the work done pushing the mass is the same.

$\endgroup$
2
$\begingroup$

The unit of joule is:

$$1~~J=1~~N\,m=1~W\,s=1\frac{kg\,m^2}{s^2}$$

Example 1

with:

$$F=m\,g~,m=\frac{F}{g}$$

Thus:

one joule is the energy that you need to raise a mass of 0.102 [kg] from zero to 1 meter.

Example 2

with: $$F=m\,a$$ and $$v(t)=b\,t~\Rightarrow x(t)=\frac{b}{2}\,t^2$$

$\Rightarrow$

$$F=m\,a=m\,\frac{dv}{dt}=m\,b\\ W=F\,x=m\,b\,\frac{b}{2}\,t^2$$

Thus:

to accelerate a body of 2 kg mass, from zero velocity to one $~[\frac ms]$ you need the energy of one joule.

To get one joule you can choose another mass and velocity or example 1 other mass and distance.

$\endgroup$
0
1
$\begingroup$

If you have a 1kg object and a force of 1N, the object will move 1 meter in about 1.4142 seconds. The 1kg object will now be moving at about 1.4142 meters per second. Note that after 1 second the object will not have travelled 1 meter yet, but it will be travelling at 1 meter per second. By the time it actually travels 1 meter, it is going about 1.4142 meters per second.

If you have a 10kg object and a force of 1N, the acceleration will be 0.1 m/s/s. It will take about 4.4721 seconds before the object has moved a meter. The object will have a velocity of about 0.44721 meters per second.

https://www.calculatorsoup.com/calculators/physics/displacement_v_a_t.php

The kinetic energy imparted is E = 1/2 * m * v^2, where E is in joules, m is in kilograms, and v is in meters per second. If we plug in the values above then we get 1 joule for the kinetic energy of both objects.

$\endgroup$
1
$\begingroup$

In both the examples, you push the same force of one newton on two different masses. By definition, work is the measure of energy transfer that occurs when an object is moved over a distance by an external force ($W=Fd cos \theta$). In both the examples in your question, the net forces and the total dicplacements are the same.

$\endgroup$
1
$\begingroup$

You have to also consider the acceleration of the body subjected to the 1 newton force.

In the case of a body of relatively little mass at the end of the one meter push its velocity would be relatively high. Instead by pushing a body with relatively high mass its final velocity would be rather small.

If the reasoning above is still not enough to make you understand with the definition of work has no need to specify the mass of the object being pushed keep in mind that work is essentially a measure of the energy transmitted by a force, and in fact the kinetic energy of a body depends not only on its mass but of course also on its velocity.

$\endgroup$
2
  • 1
    $\begingroup$ „the acceleration of the body subjected to the 1 Newton force“ mass multiply by the acceleration equal Newton $\endgroup$ – Eli May 11 at 14:29
  • $\begingroup$ The last sentence/paragraph is practically incomprehensible due to its length, missing punctuation, and other problems. Can you fix it? $\endgroup$ – Peter Mortensen May 12 at 10:07
1
$\begingroup$

The question asked for the definition of a joule. Let’s state, and restate, the elementary concepts regarding this definition...

One joule is equivalent to the energy expended, or work done, by a force of one newton accelerating a one-kilogram mass through a distance of one meter in one second. Essentially, a one-newton force, acting on a one-kilogram mass, will accelerate that 1-kg mass through one meter in one second. One newton is the force, one kilogram is the mass, and one meter is the distance moved in one second time; one joule is the energy expended, or work done, in this action. This energy expenditure is equal to exactly one watt.

Work occurs through a force acting over a distance. Energy is expended. Therefore, one joule is the work done, or energy expended, when a force of one newton acts through a distance of one meter. The energy required to complete this action is exactly one watt. If this energy expenditure happens in one second, then it is defined as one joule/second or one watt-second, or one watt/second. These units are typically given as J/s and W-s, or W/s. Also note the outcome: our 1-kg mass has a velocity of one meter per second, and if the one newton force continues, the rate of acceleration will continue at 1 meter per second per second, which is the same as one meter per second squared.

Consequently, a force of one newton acting on a 100 kilogram mass will accelerate that 100 kilogram mass through a distance of one centimeter in one second. The energy expenditure in doing so will still be equal to one joule. Note, however, in this case the mass is 100 times as great as the previously stated 1 kg, yet the distance moved is one one-hundredth of that previously stated distance of 1 m. Everything is in proportion. The energy expenditure, one joule, or one watt, is the same.

We need to also understand that concepts like heavy are related to weight, not mass. When we are talking about mass in the essence of definitions in physics we are talking about actions imparted on objects having mass (not weight) in a frame of reference without gravity. In such a frame, we can sense mass through momentum; if the object is hard to move, it is massive.

Or, if gravity is present, we are talking about the force of gravity acting on objects having mass where such objects are accelerated by gravity in free fall, or where such objects are stationary and exert a force on a stationary, horizontal platform (i.e. the force of weight, if massive they are heavy), or where objects are accelerated by gravity along a frictionless inclined plane.

Students in elementary physics classes are asked to make measurements of force, acceleration, and mass, using the classical definitions given above. The tools used in these determinations are an air table with a measured metric grid, an air rail with an attendant metric scale, a kilogram-mass air puck along with others of various masses, a spring scale, a timed-repeater strobe, and a camera for time exposures of strobe-lit experiments on the air table or rail, or for an object in free fall.

Aspects of force, mass, and acceleration, such as the resultant velocity an object gains, and distance an object is displaced, are presented to the students through the use of the differential and integral calculus, and the analysis of their measurements. These realizations are essential to understanding the concepts of potential and kinetic energy and their relation to force, mass, and acceleration.

Classic definitions of joule, a unit of work, can be found in any physics text, or in the following reference -

Sims, Frank, Engineering Formulas: Conversions, Definitions, and Tables. Industrial Press, Inc., New York, 1996. 386 pp.

$\endgroup$
1
  • 1
    $\begingroup$ Re "This energy expenditure is equal to exactly one watt." and "The energy required to complete this action is exactly one watt.": No. Power (unit watt) is the amount of energy transferred or converted per unit time. Power is not the same as energy. (Without the units, they may numerically be the same for a particular situation, but energy and power are two different things.) $\endgroup$ – Peter Mortensen May 12 at 10:34
1
$\begingroup$

Intuitively, it would seem that moving a heavy object would require more work than moving a light one. That's because your intuition involves real world things like friction, where moving a 1000 kg packing case is much harder than moving a 1 kg parcel.

Now put both objects on a frictionless surface. Not just low friction like an ice-rink, but something really frictionless, like an air-table. Now when you apply 1 Newton to either, they start to move. Once they've moved 1 metre under that force, you've done one Joule or work on them, and each has acquired one Joule of kinetic energy as a result. The 1kg package will be moving much faster now than the 1000 kg crate as it will have accelerated more quickly, and you'll have delivered that one Joule in far less time.

Even in this frictionless world, our intuition still plays false. 'Move that object' usually defaults to move it in a certain time, or at a 'reasonable' speed. Faced with a 1000 kg crate on an air-table, we'd be very tempted to push harder, to get it to accelerate quickly. That's how my internal intuition feels to me anyhow.

The one Joule only cares about the distance moved and the force applied, not how long it took, or how quickly it accelerated.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.