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What will be value of work in following cases.

I move a box on a rough floor in a straight line for a distance 'd' from A to B then B to A back.

1- Is work done by me= 0? as displacement=0 (is that correct? Or should it be +2Fd)

2- work done by Kinetic friction= -2fd. (This one is already provided in books)

3- I believe forces applied by me and friction can be same or different in magnitude therefore work done by me and friction can be same or different in magnitude. Or Is it necessary that work done will be equal in magnitude?

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    $\begingroup$ The idea of work is confusing unless you consider the integral formula for work: $W = \int_{\text{path}} \vec{F}\cdot d\vec{r}$. $\endgroup$ Aug 28, 2022 at 16:41

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Work is defined as dot product of force vector applied and the displacement vector caused due to that force.

So for very small displacement $\vec{ds}$ caused due to some force $\vec{F}$,
the small amount of work done will be: $$dW=\vec{F}.\vec{ds}$$ So total work done over a path (say A to B) will be: $$W=\int_A^B \vec{F}.\vec{ds}$$

In your question, even if displacement is zero but you have done positive work in both trips i.e. A to B then B to A. This is because in both the trips displacement is in same direction as force applied, so the dot product is positive so the work done.

Note that if there was no friction then work done will be zero in both the trips and also overall. While going from A to B you first apply a force causing block to move in forward direction; here you are doing positive work and Kinetic energy of block is increasing (Work energy theorem). But you also have to stop at B and for stopping you will have to apply a force in opposite direction of the motion. Work done by this force should be negative but equal in magnitude with the previous mentioned force. This is because, when the block stops at B its kinetic energy is zero so the net work done must be zero during the trip. Some goes for trip B to A.

If we consider friction also then things will go a bit different. While considering friction it is believed that applied force is just enough to overcome maximum static friction ($\mu_s N$) and thus the block moves very very slowly (zero kinetic energy). In this case we don't need to apply another external force to stop the block since equilibrium is maintained at all points (due to very very slow movement). So work done by external force (applied by you) and work done by kinetic friction adds up to zero. Same goes for B to A trip.
Thus overall work done is zero but work done by you is positive (2Fd) and work done by friction is same but with negative sign (-2Fd).


Note that in this case block does not have any velocity while moving (very slow movement); but if applied force is large enough than kinetic friction then block will gain velocity and it may occur that you have to apply an another external force to stop the block on time i.e. at B.

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  • $\begingroup$ Thank you for detailed answer. $\endgroup$
    – Level1
    Aug 30, 2022 at 15:05
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  1. No. To move a body in presence of friction (assuming constant velocity), you need to do a force in the direction of motion, both when you move to the right (force and displacement pointing to the right) and you move to the left (force and displacement pointing to the left). So the work you do is $W = 2 F d$

  2. The work done by friction is exactly the opposite $W_{friction} = -2 F d$, since $F_{friction} = -F$.

  3. At the end, the work done by you is the opposite of the work done by friction must if the kinetic energy of the initial and final time is the same $\Delta K = 0$, and no other force produces work on the system. This is true since the theorem of kinetic energy reads

    $\Delta K = W^{tot} = W + W_{friction}$.

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  • $\begingroup$ Thank you for a brief answer. $\endgroup$
    – Level1
    Aug 30, 2022 at 15:58
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    $\begingroup$ You're welcome. Enjoy $\endgroup$
    – basics
    Aug 30, 2022 at 16:12

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