Does the work done by a person equal the work done on the object in this situation?

Question

If you applied a constant force over a floor that has friction on an object, would the work done by the person equal to the work done on the object? Assume that the floor is flat and that the object is being pushed to the right.

For example, if I push an object with 300 N and the work done by me is 200 J, would the work done on the object also be 200 J?

I would say yes because they both share the same forces. These forces would be, if you were to draw them on a force diagram, the force of gravity, friction, the normal force and the force being applied.

In other words, the person and the object would have the same force diagram.

Answer 1

The work done by the person equals the work done on the object by the person, but it is not equal to total work done on the object, because friction forces do work on it as well.

Answer 2

The answer is actually no.

The kinetic energy of the object is not changing, hence the net work on the object is equal to 0 J. You are doing positive work on the object, but friction is doing negative work; taking energy. So you can still be doing work on the object, but the net work on the object, which is what changes the kinetic energy, can be 0. This is similar to how an object can be mechanical equilibrium even if there are forces acting on it. I hope this helped answer your question.

Answer 3

case Ist: work done by object

On object there are four forces- $mg$, normal force, force exerted by man and frictional force. Normal contact force and $mg$ are perpendicular to the direction of the force exerted by the man and the frictional force hence not going to include them but in x-direction force by man and the friction are opposite hence the net force is $$F-\mu N=F_{net}=ma\ldots(1)$$ $m=$mass of that object and $N=mg$ and now suppose the object is displaced to a distance of $s$ meters in the direction of $F$, hence equation (1) is $$Fs\times cos(0^o)-\mu Ns\times cos(180^o)=F_{net}s\times cos(0^o)\implies Fs+\mu Ns=F_{net}s =W \ldots (2)$$

case IInd: work done by man

if you are putting a force $F$ on the object then the object will exert an equal and opposite force $-F$ and $\mu N$, man's displacement is $s$ in the opposite direction so the work done by the man is $$W=-F\times s\times cos(180^o)+\mu Ns\times cos(180)=Fs-\mu Ns \ldots (3)$$. You will question that why i haven't included the frictional force on the man, its because of the fact that friction is helping the man to put force on the box.

so from equation (2) and (3) we can see that man has done less work

Answer 4

To deal with the question we need a definition for work.   The truest definition is that work is the transfer of energy from one system to another by a force exerted through a distance.   It is very instructive to apply this definition directly, because it leads to a clear analysis in every situation, whereas the work concept can be confusing and is often applied incorrectly.

In the case of the man pushing the block, the energy from him is transferred to the system comprising the block and the floor.  It starts as chemical energy in the man and ends up as kinetic energy in the block plus thermal energy in the block, plus thermal energy in the floor.   So the work done by him is equal to the work done on the system consisting of the block and the floor.

We can't say that the friction force transfers energy out of the block without implying that all of the energy transferred from the man first goes into the block and gets briefly stored there somehow.   Otherwise, what energy would the friction be transferring out of it?   Therefore, we can't correctly say that the friction force does work on the block unless we also say that the work done by the man is equal to the work done on the block.   Interesting, isn't it?