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So, in defining work as force * distance, how is mass applied? Is it just that it's $1kg$ multiplied in normally, but that's typically ignored / not shown?

..so:

$2kg \cdot 1n \cdot 1m$ = $2J$

..or is the answer for $2kg$ still somehow $1J$, and I'm missing something?

@knzhou points out that mass is irrelevant -- however, it took me a while (and a different conversation) to sort out why.

Edit: In retrospect, the question would probably have been better phrased as *"Why isn't mass relevant in calculating work?"

The thought experiment I had in my head was this:

Two bricks, a 1kg brick and a 2kg brick, are floating in space. A 1N force is applied to each one until it moves 1 meter. It takes longer for the 2kg brick to reach this point, and I was thinking that since the force was applied for a longer time, it amounted to an increase in kinetic energy (and thus it must have recieved more work than the 1kg brick). However, this isn't the case -- it has more momentum, but the work received is the same, reflected in that the 2kg brick has a lower velocity after having travelled the meter.

In the end, I was conflating work and momentum, and overlooking the final velocity of each brick. Thanks for the answers, and I apologize that my question wasn't well-structured.

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closed as unclear what you're asking by Diracology, Gert, innisfree, CuriousOne, honeste_vivere Jul 28 '16 at 12:00

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    $\begingroup$ Maybe it's easier to understand if you imagine the force being used to lift something. Lifting a 2 kg object by one meter indeed takes twice as much work as lifting a 1 kg object the same amount. But that's already accounted for in the formula because the force has to be twice as much. Multiplying by mass again is wrong. $\endgroup$ – knzhou Jul 28 '16 at 0:10
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    $\begingroup$ @knzhou You comment is a better answer than either of the posts below. Why not make it an answer. $\endgroup$ – dmckee Jul 28 '16 at 0:29
  • $\begingroup$ @knzhou was the closest to clarifying my misconception (I sorted it out in a different conversation). $\endgroup$ – Mr. B Jul 28 '16 at 5:31
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There are two quantities that are relevant to this discussion: kinetic energy and momentum.

The change in an object's kinetic energy is equal to the work done, which is forces times distance. $$K_2 - K_1 = W = F\cdot (x_2 - x_1)$$ It does not matter whether the object is a bowling ball or a tennis ball. If you apply 1 newton of force over 1 meter, the object will have 1 joule of kinetic energy.

The change in an object's momentum is equal to the impulse applied, which is force times time. $$p_2 - p_1 = Imp = F\cdot (t_2 - t_1)$$ Again, the mass of the object doesn't matter. If you apply 1 newton of force over 1 second, the object will have 1 newton-second (or, 1 kilogram-meter/second) of momentum. There's no name for units of momentum like joule for energy.

Kinetic energy and momentum are related, and it takes mass to express that relationship. $$K = \frac{mv^2}{2} = \frac{m^2v^2}{2m} = \frac{p^2}{2m}$$ If two objects have the same momentum, the one with greater mass has less kinetic energy. The fact that they have the same momentum means that they were pushed with the same force over the same amount of time. The greater mass would have moved less, so it received less work, thus less kinetic energy.

Solving for $p$: $$p = \sqrt{2mK}$$

If the two objects have the same kinetic energy, the one with greater mass has more momentum. The two objects having the same kinetic energy means they were pushed with the same force through the same distance. The greater mass would have been pushed for a longer time due to its slower acceleration, so it received a larger impulse.

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Mass is accounted for already in the force.

$F = m \cdot a$

Then work formula could also be

$W = m \cdot a \cdot d$

where the mass $m~ [kg]$ is accelerated at $a ~[ ^{m}/_{s^2}]$ over the distance $d ~[m]$ which is equal to $W ~[N \cdot m]$ or $[\frac{kg \cdot m}{s^2} \cdot m]$

Always look at units they are probably the most important thing to use to make sense of things and see how things relate.

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In SI units: $1~\mathrm{N} ~\mathrm{m}=\mathrm{1~kg~ m^2 ~s^{-2}} $ and $1~\mathrm{J}=\mathrm{1~kg~ m^2 ~s^{-2}} $. So the units Newtonmeter and Joule are the same in SI-Units and there dimensions are actually equal too. This is a fact and just a matter of definition.

Work along a curve $C$ is $W=\int_C \vec{F} \cdot d\vec{s}$. So in the simplest case where the force is constant and parallel to the curve, the work along it is Force times length of the curve. So Newton times meter.

And to work in the gravitational field: To lift an object of mass 1kg 1 meter one would need to perform $W=F_G*x=1\mathrm{kg}*g*1\mathrm{m}=1\mathrm{kg}*9.81\mathrm{m s^{-2} *m} = \mathrm{9.81~kg~ m^2 ~s^{-2}}$. Where g is the gravitational acceleration not the gravitational force. Mind the difference between force and acceleration: $F=m*a$ for constant masses.

You do not apply a force to a mass and then integrate to get work that is just wrong. If you bring in "force to a mass over a time" you do not compute work; in that case you solve an equation of motion: $m\frac{d\vec{x}}{dt^2}=\vec{F}(x)$. Another quantity in that context would be energy which has also the SI unit Joule. The kinetik energy of a classical point-like mass of mass M and momentum p would be: $E_{kin}=\frac{1}{2m} \vec{p}^2$ with $\vec{p}=m\frac{d\vec{x}}{dt}=m \vec{v}$.

This was all a bit mathematical but if you search for Work, Force, Energy, ... you can find more literal definitions and explanations but you need to keep context of what you want to compute/know. Only because something has the same unit does not mean it is the same.

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It is true that $Work = Force * Distance$. Picture a mass, $M$, in empty space away from all gravitational fields. In other words, it's not moving at all, it's not accelerating at all.

Then, an acceleration, $g$, suddenly appears and now we have the situation where this acceleration results in a force being applied to the mass. By one of Newton's basic laws, this $Force= M*g$. This force is constant. It doesn't change in our case. And it's this acceleration, from "who knows where", that's causing this force relationship.

This force causes the mass to accelerate from $Velocity = 0$ because it wasn't moving to begin with. The mass changes its position as it follows a straight line. Its position changes from $X_{initial}$ to $X_{final}$ along the straight line that follows the exact direction of the force. This force has caused our mass to undergo acceleration giving the mass "punch" or "energy" that it didn't have before. This energy is quantified as $Work = M*g*(X_{final}-X_{initial})$. Mass is usually in kilograms. $g$ is acceleration and is usually in meters per second squared. The $X's$ are usually in meters. You can see how mass figures into the equation.

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    $\begingroup$ Thank you for your response! The part I don't understand is the discrepancy between Work=Force∗Distance vs Work=M∗g∗(Xfinal−Xinitial) (or, Work = Mass * Force * Distance). $\endgroup$ – Mr. B Jul 28 '16 at 0:31
  • $\begingroup$ So mass is just not mentioned, normally, even though it's a part of the equation? $\endgroup$ – Mr. B Jul 28 '16 at 0:40
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    $\begingroup$ No. Mass it not part of the equation. Force is. But if the motion is lifting against gravity then the force is proportional to mass. But to be clear: the mass enters because you are counteracting gravity and not because it is in any way built into the notion of work. $\endgroup$ – dmckee Jul 28 '16 at 0:44
  • $\begingroup$ @Mr. B Force equals mass times acceleration. That's where I got Force = M*g from. $\endgroup$ – Inquisitive Jul 28 '16 at 23:50
  • $\begingroup$ @dmckee If a gravitational force is accelerating a mass at "g", then $Force = M*g$ and that force caused by "$g$" is doing work on the mass. That's how the mass gets its velocity. It's described by a "free body diagram" of a mass with one force, $M*g$, acting on it and accelerating it from $V=0$ to $V=$something greater. $\endgroup$ – Inquisitive Jul 28 '16 at 23:55
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Force is mass times acceleration. From this you can get work as mass times acceleration times distance.

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  • $\begingroup$ It is often useful to compute the work done by individual forces even in the absence of acceleration. $\endgroup$ – dmckee Jul 29 '16 at 0:26

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