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$$W=\left\Vert\vec{F}\right\Vert \left\Vert\Delta\vec{r}\right\Vert \cos{\theta}$$ A greater magnitude force will have a greater influence than a smaller magnitude one when they have the same $\theta \neq \frac{\pi}{2}$. Also, when the angle between the force vector and the displacement vector($\theta$) is smaller, the influence of the force on the motion of the object is greater.
But, the force has nothing to do with the magnitude of the displacement, since forces are generators of accelerations not displacements (The force may be great and $\theta$ may be small but the magnitude of the displacement is still small), so, why displacement is invovled in the work difinition?

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  • $\begingroup$ I am having a hard time coming up with an explanation other than "because if it does things work out correctly in mechanics". I feel the correct answer would provide some historical context into the concept of work. $\endgroup$ – ja72 Dec 1 '18 at 23:13
  • $\begingroup$ Push a wall for a while, you might get tired, but have you done any work? (Rigurously: has your force done any work on the wall?) Nothing has changed. $\endgroup$ – FGSUZ Dec 2 '18 at 1:01
  • $\begingroup$ My guess: everyone will agree that work requires energy, so the units that any physics definition would implement, should strive to have work end up as an energy entity. Dimensional analysis indicates that work has units of Joules, or Newton-meters, where Newton-meters is seen to be force multiplied by distance. $\endgroup$ – David White Dec 2 '18 at 1:31
  • $\begingroup$ "...forces are generators of accelerations not displacements" (??). Acceleration is the rate of change of velocity. Velocity is the rate of change of displacement. Ergo if forces "generate" accelerations they also generate displacements and that's how they perform work. $\endgroup$ – Bob D Dec 3 '18 at 2:22
  • $\begingroup$ @FGSUZ I think if you look at the muscles, they are contracting and expanding during this process and that is where work is done. $\endgroup$ – jim Dec 14 '18 at 12:49
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Work is a measure of how much energy one adds to a mechanical system. Consider an object initially at rest on a frictionless surface, such as an air hockey puck on an air hockey table. Furthermore, consider pushing with a constant horizontal force $F$ starting at time $t_0$. For simplicity, we are assuming here that $\theta=0$. If one applies the force until the air hockey puck has traveled a distance $x$, the final kinetic energy after the force is applied will be $E=Fx$. Specifically, the larger the distance over which the force is applied, the larger the kinetic energy will be of the hockey puck once the force application period is complete.

Consider two limits. If the hockey puck is pushed over an infinite distance, its velocity will approach infinity. If the hockey puck is pushed over zero distance, the velocity will be zero.

Work is a measure of energy. Can you see now why the displacement is relevant for determining the work?

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    $\begingroup$ Just a suggestion, but you might want to say "adds to or removes from a mechanical system", since work can be positive or negative. I would also not restrict it to mechanical systems (though I realize the subject is mechanics), since it also applies to work in an electrical field. $\endgroup$ – Bob D Dec 3 '18 at 11:54
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The most basic definition of work is as follows: work is the amount of energy a force transfers between objects. The "force multiplied by distance" definition of work follows as a direct consequence of this definition, because it turns out that it's impossible to transfer energy to an object without either moving the object itself or moving around its internal degrees of freedom.

If you transfer energy to a block, then that block has to gain some kind of energy:

  • If it gains kinetic energy, then it starts to move faster, and the notion that a force applied over a distance increases energy makes sense.
  • If it gains some kind of potential energy, then the configuration of the objects involved must change, which implies a change in position.
  • If it gains thermal energy, then the internal degrees of freedom (for example, the atoms in the lattice of a solid) gain kinetic energy, so they move faster under the influence of various resultant forces of the initial force propagating through the lattice (and in this case, they change their rms displacement from equilibrium in the lattice sites).
  • If it is compressed or stretched, then the atoms in the lattice gain electrical potential energy, which involves a deformation of the lattice and hence a change in position. In all cases, displacement is necessary for the block to gain energy, which is why the "force multiplied by displacement" definition makes sense.

Even dissipative forces still follow this definition:

  • If it loses energy to air resistance, then it is increasing the kinetic energy of the internal degrees of freedom of the air (in this case, the molecules that collide with the block.
  • If it loses energy to electromagnetic radiation, then that radiation is generated by decreasing the kinetic energy of the internal degrees of freedom (the charges in the lattice of the block).

Hence, if there is no displacement, either in the object or in the internal degrees of freedom of the object, then there can be no energy transfer and therefore no work done.

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