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We define work done as force applied multiplied by displacement. Let's say a field applies a force $F_1$ on a body $b$. Let the distance between point $A$ and point $B$ be $d$

Then work done in moving the body from point $A$ to point $B$ (against the field) will depend on the force we applied to move the body from A to B. Let the force we applied be $F_2$. Since $F_1$ is being applied on the body, we should need to apply a force greater than $F_1$ if we want to move the body, i.e. $F_1 < F_2$. So work done $W$ will be $W = F_2 \cdot d$

But in my book it is given that ⠀$W = -F_1 \cdot d$ ‎‎‎‎‎‎‎⠀i.e.⠀ $F_2 = -F_1$

So am i wrong in assuming $F_1 < F_2$, or do we consider $F_2$ to be extremely close to $F_1$ and let it equal $F_1$, or something else

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2 Answers 2

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You need to look at the average force applied to the body between points A and B.

In order to initiate movement of the body initially at rest from point A against the force of the field, you need to apply a force greater than the force of the field to accelerate the body. Then in order to bring the body back to rest at point B you need to reduce the applied force below the force of the field to decelerate the body. This makes the average applied force equal to the force of the field and the change in kinetic energy of the body between A and B zero.

The work done by the external force is positive since it is in the same direction as the displacement of the body. Positive work transfers energy to the body. The work done by the field force is negative since its force is opposite the direction of the displacement of the body. Negative work takes energy away from the body.

In this case, the negative work by the field takes the energy given the body by the positive work of the external force and stores it as potential energy in the system of the body and the source of the field (mass for gravitational potential energy, charge for electrical potential energy, etc.)

Hope this helps.

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    $\begingroup$ btw can you link me to a book or some text online which states this $\endgroup$ Apr 19, 2022 at 15:46
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    $\begingroup$ @Gurneetsingh What part of my answer are you referring to? $\endgroup$
    – Bob D
    Apr 19, 2022 at 16:07
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    $\begingroup$ nvm i understood it $\endgroup$ Apr 19, 2022 at 16:15
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If $F_1$ were the only force acting on the body, the body would accelerate in the direction of the force $F_1$. If you apply force $F_2$ that is equal in magnitude and opposite in direction to force $F_1$, by first Newton's law of motion the body will be in equilibrium. The equilibrium means that the body will keep moving at constant velocity, which can (but does not have to) be zero.

Note that both velocity and distance depend on reference frame. In one inertial reference frame the body might seem to be at rest, while in some other it seems to be moving at some constant velocity. This is why first Newton's law mentions "equilibrium" rather than "rest" or something similar.

Obviously, if $F_1$ and $F_2$ are the only two forces acting on the body, when $F_2 = -F_1$ then the net work done on the object is zero. By work-energy theorem

$$\Delta K = W$$

this means that there is no change in its kinetic energy, i.e. velocity is constant as already discussed above.

Since $F_1$ is being applied on the body, we should need to apply a force greater than $F_1$ if we want to move the body

Body can move (at constant velocity) even if you apply $F_2 = -F_1$. Once you start applying the force $F_2$, the net acceleration will be zero and the body will move keep moving at the velocity which was at the moment you started applying force $F_2$.

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