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This is a question about the nature of work, in classical mechanics.

An object has 5N applied to the right. it gets displaced 5m to the right

An object has 20N applied to the right, but friction force is 15N to the left. it gets displaced 5m to the right.

Which case has more work been done?

By intuition, it should be the first, since more physical exertion is needed for the same displacement, hence more work.

However, work is also the same as change in energy, and in both cases the change in kinetic energy is about the same.

More overall, what is the formula for work? W = (F)(s)cos(theta), or more accurately, W = (Fnet)(s)cos(theta)?

This is a very confusing matter for me.

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  • $\begingroup$ Work done is by net force which moves the block, in (b) case net force is $20N-15N=5N$. Hence (b) total work is same as in (a) case. $\endgroup$ Dec 9, 2023 at 10:37
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    $\begingroup$ You are dealing with work done by one force of 20 N, work done by a frictional force of 15 N, and net work (the work done by the net force). The fact that there are three possible answers to this problem is probably one of the sources (perhaps the main source) of your confusion. You need to specify which work you are interested in calculating. $\endgroup$ Dec 10, 2023 at 0:46

4 Answers 4

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System - object

Case $1$
External force $5\,\rm N$ to right.
Work done by external force on object when displaced $5\,\rm m$ to right = $5 \times 5 = 25 \,\rm J$.

Case $2$
Two external forces, $20\,\rm N$ to right, and $15\,\rm N$ to the left which is the same as $-15\,\rm N$ to the right.

Work done by external $20\,\rm N$ force on object when displaced $5\,\rm m$ to right = $20 \times 5 = 100 \,\rm J$.
Work done by external $-15\,\rm N$ force on object when displaced $5\,\rm m$ to right = $-15 \times 5 = -75 \,\rm J$.
Net work done on the object by the two external forces is $100-75=25 \,\rm J$, the same as in case $1$.

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  • $\begingroup$ does this suggest that power can have three 'outcomes' in case three - depending on considering the power of the 20N force, -15N force and the net work? $\endgroup$
    – Jay Chen
    Dec 9, 2023 at 10:26
  • $\begingroup$ Case 3 if one is considering the net force is the same as case 1. $\endgroup$
    – Farcher
    Dec 9, 2023 at 14:07
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Work is the transfer of energy. Determining which case is "more work" depends on whose work you are talking about.

Let's say I apply the forces to the right that push the object. To calculate the work done by me I consider: $$ W_\mathrm{done\, by\, me} = \vec{F}_\mathrm{applied\, by\, me} \cdot \Delta\vec{x}.$$

This same idea applies to any force.

If the parallel components of $\vec{F}$ and $\Delta\vec{x}$ both point the same direction (as with my force), then the dot-product will give a positive result. The force added energy to the object.

If the parallel components of $\vec{F}$ and $\Delta\vec{x}$ point opposite directions (as with friction), then the dot-product will give a negative result. The force removed energy from the object.

Case 1:

  • I did $\vec{F}_\mathrm{me}\cdot\Delta\vec{x} = +25$ J of work. I transferred $25$ J of my energy to the object.
  • No energy was lost to the environment.
  • The object experienced a net of $25$ J of work. It gained $25$ J of energy during the process.

Case 2:

  • I did $\vec{F}_\mathrm{me}\cdot\Delta\vec{x} = +100\,\mathrm{J}$ of work. I transferred $100$ J of my energy to the object.
  • Friction did $\vec{f}\cdot\Delta\vec{x} = -75$ J of work. $75$ J of the object's energy was removed from it and transferred into the environment through friction.
  • The object experienced a net of $(\vec{F}_\mathrm{me} + \vec{f})\cdot \Delta\vec{x} = 25$ J of work. The object gained a net of $25$ J of energy during the process.

So it depends on whether you are asking about the work done by me or the energy gained by the object. In each case the object gained the same energy, but I did more work in the second case because some of the energy I transferred was lost to friction.

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When talking about work, you need some surnames:

Work... [exerted on which body][by which force] and [along which trajectory]

If you are talking about the work done on the object along those 5 meters (assuming it is on a straight line), you still need to clarify the work you are calculating:

  • The work done by the force $F$ will be $W_F=F\cdot(5\ m)\cdot cos(0º)$
  • The total work done on the body will be the sum of the previous work plus the work done by friction
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  • $\begingroup$ So in the second example I gave, the work done by friction on the object, and the work of the force done on the object is equal to the total work done on the object? $\endgroup$
    – Jay Chen
    Dec 9, 2023 at 9:57
  • $\begingroup$ Not quite, the TOTAL work (net work) on the object is the sum of all individual works done by each individual force $\endgroup$
    – FGSUZ
    Dec 9, 2023 at 14:34
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You can't just ask the formula for work without specifying You can ask the formula for work done by the force on the object, which would be $Fscos\theta$, where F is the magnitude of said force. This work can be calculated for both the forces respectively, and their sum is the net work done on the object by both the forces.

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