# Mechanics: what does work done against resistance really mean?

Does ''work done against resistance'' mean the force needed to overcome friction i.e. the magnitude of the frictional force itself x distance? or does it mean the work done by the net force i.e. (the driving force - the resistance force) x distance? Could you please give a clear and precise definition?

• TLDR: If you exert consant force on a thing while the thing moves, then the dot product of the force and the displacement of the thing tells you how much work you did on it (or, depending on the sign, how much work it did on you.) Feb 15, 2021 at 15:26
• To be completely accurate, work is done by a force, not an object. The work done by frictional force is the dot product of frictional force and the displacement of the point of contact. It is usually negative (friction depleting energy) but can also be positive (friction adding energy) Feb 15, 2021 at 20:09
• thanks for answering this question. Feb 16, 2021 at 12:19

Take the example of pushing a box on a floor. There are two friction forces to overcome, static friction (which does not involve work) and kinetic friction (which does involve work).

When you initially try and push the box it doesn't move until the force you apply exceeds the maximum static friction force that opposes your applied force. The maximum static friction force is

$$F_{s-max}=\mu_{s}mg$$

Where $$\mu_{s}$$ is the coefficient of static friction.

At that point the box breaks free and starts to slide. Now the force that opposes you is the kinetic friction force, which is

$$F_{k}=\mu_{k}mg$$

Where $$\mu_{k}$$ is the coefficient of kinetic friction. Generally the coefficient of kinetic friction is less than the coefficient of static friction so that the force you needed to apply to keep the box moving is less than the force you needed to apply to start it moving.

Now, two things can happen.

Since the kinetic friction force is less than the static friction force, if you continue to apply the force that started to make the box slide it will be greater than the kinetic friction force so the net force will cause the box to accelerate. On the other hand, if you reduce your force to exactly equal the kinetic friction force, or

$$F_{applied}=\mu_{k}mg$$

then the net force will be zero and the box will slide at constant velocity. Now the work you do sliding the box, say a distance $$d$$, will be

$$W_{you}=F_{applied}d$$

At the same time friction does an equal amount of negative work (because its force is opposite the displacement of the box), or

$$W_{friction}=-\mu_{k}mgd$$

for a total net work of

$$F_{applied}-\mu_{k}mgd=0$$

The end result is friction takes the work you do and dissipates as heat at the sliding surfaces.

Hope this helps.

• thank you, i am studying maths, not physics but i was just concerned about what the meaning of the sentence is. not why fmax = UR. i just want to know what the sentence ''work done against resistance'' literally means. Feb 15, 2021 at 15:24
• From my example, it means if you want to slide the box along the floor with friction, you have to do apply a force at least equal to the friction resistance force. Then when you move the box the work you do is "work is done against resistance", where "resistance" means against the friction force. Feb 15, 2021 at 16:47
• what if you apply a force greater than the resistance force? then the net force is the pulling force - the resistance force. so then, the net work done is the work done by the pulling force minus the work done against resistance. that's what i think you should have said to be more clear but thanks for your answer anyway. Feb 19, 2021 at 17:55