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Is the force $F$ the resultant force on the object, or exactly the force needed to move the object?

e.g.: I might 'need' only 1 N to move an object over 1 metre, but if I apply a 10 N force to that object over 1 metre, am I still doing the same work as if I applied a 1 N force over the same 1 metre distance?

Or, if I use a resultant force of, say, 100 000 N upwards, to lift an object of 1 kg, 1 metre above the ground, what is the work being done?

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  • $\begingroup$ You're confusing terminology. The resultant force is the net total of all forces applied. What is "The force needed to move an object"? Absent friction, even the smallest force will make even the largest object accelerate. $\endgroup$ Apr 28, 2017 at 1:39
  • $\begingroup$ This is indeed confusing me. How should I answer the two examples I gave? Getting answers to various examples should help me understand what is meant by the equation of work. $\endgroup$
    – Stephen
    Apr 28, 2017 at 1:46
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    $\begingroup$ Can the two people who voted to close this as homework explain to me how they view this as a homework problem? OP seems to have a confusion about forces & work and doesn't seem to be related to any particular problem. $\endgroup$
    – Kyle Kanos
    Apr 28, 2017 at 10:08
  • $\begingroup$ @KyleKanos Indeed. This was not a homework at all. I'm currently studying math, but I'm also very interested in physics, which makes me study it as a mere hobby when I have time. This means that my knowledge of physics is somewhat scattered all over the place. I’m familiar with some advanced topics, while some basic ones still remain elusive – which is the reason behind this thread. $\endgroup$
    – Stephen
    Apr 29, 2017 at 0:13
  • $\begingroup$ @KyleKanos Nice to see this comment. $\endgroup$
    – peterh
    Apr 29, 2017 at 20:47

2 Answers 2

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Work is done by a thing applying a force. Whatever force you name defines the work done.

If you apply 10N to a an object over the distance of 1 metre, you did 10N*1m=10J of work. If you wish to phase it as there being a "needed" portion of the force and an "unneeded" portion, you could say that the "needed" portion of the force did 1N*1m=1J of work.

The latter is not typically useful, but I use it to point out how you can divide up the forces in any way that you find meaningful, and we can talk about the work done by each and every one of those forces.

You can even play games such as saying that that 10N force was actually 100N in the direction of motion and 90N against it. There's nothing that prevents it, though it might be awkward. In such a case, you would say the thing applying the 100N did 100N*1m= 100J of work on the object, and the thing applying the 90N in the other direction did -90*1m=-90J of work. As you can see, if you add up the 100J of work done by one force, and the -90J done by the other, you still get the same 10J of work that was done in total. Any way you split it up is fine.

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  • $\begingroup$ Thanks. How would you address the second example with gravity? My textbook seems to imply that lifting an object of say 1kg, 1 m above the ground, translates into a work of 9.81J, no matter the force used to lift the object. I'm not sure how to understand this. $\endgroup$
    – Stephen
    Apr 28, 2017 at 1:51
  • $\begingroup$ That would be true if it didn't gain any velocity from start to end. If you applied 10000N to the book over the entire 1 meter, it would be traveling very fast, and that velocity means kinetic energy. Work would have gone into building up the kinetic energy. The fascinating part of this whole energy discussion is that, if the book starts at rest, and ends at rest 1m higher, it actually wont matter how it got there... you'll still have put 9.81J in (later, you can prove this to be true with calculus), but if the book gains velocity from start-to-end, it's a different story. $\endgroup$
    – Cort Ammon
    Apr 28, 2017 at 1:58
  • $\begingroup$ I see. Here's the actual question that confused me in my textbook: In 1957, the US ‘strongman’ Paul Anderson claimed an (unofficial) world record that is still unbeaten 50 years later. He lifted a platform holding a total mass of 2840 kg. In doing this, he did 280 J of work on the platform. Through what distance did he raise the platform? Give your answer to the nearest centimetre. Take g = 9.81 m/s². $\endgroup$
    – Stephen
    Apr 28, 2017 at 2:06
  • $\begingroup$ So, if this Strongman was a 'superhuman' and was for example able to lift the platform with a force of say, 1000 kN, that wouldn't change the work of 280 J he did because the platform started from rest and ended its motion at rest? $\endgroup$
    – Stephen
    Apr 28, 2017 at 2:10
  • $\begingroup$ I think I get it now. Basically, work is really the average change in energy transfer. So in the above example, some energy was used to accelerate the platform to a very high velocity, but then almost as much energy was used to slow it down to rest. The tiny difference of 280 J, in the energy used to accelerate it and the energy used to slow it down, is then the effective work done. Right? $\endgroup$
    – Stephen
    Apr 28, 2017 at 2:22
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If you use the net force, then you are calculating the net work $$ W_{net} = \int\limits_\text{path} \vec{F}_{net} \cdot \mathrm{d}\vec{s} \;.$$ This is useful because the net work appears in the work-energy theorem.

If you use some other force, say the normal force or the weight (gravitational force), \begin{align} W_N &= \int\limits_\text{path} \vec{F}_N \cdot \mathrm{d}\vec{s} \tag{normal}\\ W_g &= \int\limits_\text{path} (-mg\vec{y}) \cdot \mathrm{d}\vec{s} \tag{gravity} \;. \end{align} then you are calculating the energy transferred by the force you considered only. This is useful when bootstrapping the existence of potential energies, and leads to the whole infrastructure of energy conservation computed as a set of terms associated with individual interactions.

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  • $\begingroup$ So, in fact, there are many types of work, depending on what you choose the force to be. There isn't one single definition for the force in the formula for work (as opposed to the force in the formula F=m*a, which always uses the resultant force)? $\endgroup$
    – Stephen
    Apr 28, 2017 at 4:37

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