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I'm confused with the definition of work in physics. I know that it's defined as a product of force, displacement and a cosine of angle between them: $W = F s \cos(\alpha)$. But that means that the work of moving an object from one place to another depends on the absolute value of force we apply to it provided the displacement and the angle stay the same. But as I understood, the work equals to the total energy given away to move the object. But with weaker force and the same distance we will move the object to the finish point just with more time and the energy intuitively should stay the same. So why is that the case that weaker force corresponds to less work done? Thank you.

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No, the energy wouldn't be the same if you move an object the same distance with a weaker force. Let's see this explicitly. The energy that is given to the object during this process will be reflected in what property of the object? It will be reflected in its speed. So, if the speed of the object at the end of the process is smaller for the weaker force then we can explicitly see that the process carried out with a weaker force will, in fact, be pouring less energy into the object. So, let's see what is the speed $v$ at the end of this process if it is carried out under the influence of a force $F$. For simplicity, I will assume that the force is in the direction of the displacement.

Assuming a constant force, the acceleration of the object will be $F/m$. Now, since the object travels a distance of $s$ starting from rest, we can write $s=\frac{1}{2}\frac{F}{m}t^2$ where $t$ is the time it takes for the process to complete. As you can see, you are right that it would take a larger $t$ if we choose a smaller $F$ to cover the same distance $s$. But, now, let's compute the speed that the object will have at the end of this process. For uniformly accelerated motion, we can write $v=\frac{F}{m}t$. Expressing this $t$ in terms of $s$ from our previous formula, we can write $$v=\frac{F}{m}\sqrt\frac{2ms}{F}=\sqrt{\frac{2Fs}{m}}$$

So, indeed, the weaker the force, the smaller the final speed will be.

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  • $\begingroup$ Yes, the speed is indeed smaller, but we are pouring the smaller amount energy for a greater amount of time. And I thought that the energy corresponding to the work done from the formula shows the overall energy that we have given to the object during the period of time when we were affecting the object. In this case it seems that the overall energy's value depends also on the time the force is applied to the object. Where am I wrong here? $\endgroup$
    – DanyloL
    May 7, 2020 at 13:41
  • $\begingroup$ @DanyloL Yes, we are pouring in energy for a longer amount of time. But even after all that long amount of time, it has a smaller velocity velocity than it would have had if the force had been greater. So it's not that we are pouring in the same energy at a slower rate but it's that we are pouring in smaller energy total energy at a slower rate. $\endgroup$
    – user87745
    May 7, 2020 at 18:35
  • $\begingroup$ So, the amount of time the weaker force was applied doesn't make the total energy given by the weak force come even close to the total energy given by a strong force? $\endgroup$
    – DanyloL
    May 7, 2020 at 19:22
  • $\begingroup$ @DanyloL What do you mean come close? The exact expression is what I mentioned, i.e., for the fixed distance $s$, the final speed $v\propto \sqrt{F}$. So, given how stronger one force is compared to the other, you can calculate how bigger one speed would be compared to the other and conclude if that's close enough for you. But yes, the stronger the force, the higher the final speed would be. $\endgroup$
    – user87745
    May 7, 2020 at 20:03
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Nice question. So , I am considering a situation where we are rolling a ball up a hill. Neglect any friction and assume the hill is a perfect inclined plane making an angle (theta) with the horizontal. Let the mass of the ball be M. And let's consider rolling the ball a distance D up the hill.

To move the ball up the hill , the minimum force we need to apply on it is Mgsin(theta) (to balance gravity)

So consider we are applying a force 2Mgsin(theta) on the ball . Let's consider the energy changes which happened after rolling the ball a distance D up the hill .

The gravitational potential energy of the ball ( rather the earth - ball system) increases by some amount . The ball gains some velocity and hence gains some kinetic energy .

Now say we apply a force 4Mgsin(theta) , clearly the work done by us on the ball is doubled. But the change in gravitational potential energy is the same!

What about kinetic energy? As we applied a greater force on the ball , the net acceleration of the ball was more than the first case and hence the velocity gained by it after rolling the same distance D up the hill will be greater.

Hence , the kinetic energy gained by the ball is more than the first case . In other words , you can say that the additional work goes into the kinetic energy of the ball.

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  • $\begingroup$ I just don't get why the work done is doubling when applying a stronger force 4Mgsin(theta). We apply force for a period of time, and the stronger the force, the less time we need it to move the ball up the hill, hence the time will be less when we use stronger force, so we might not be sure about the overall energy we given away, as we were giving more energy for a less period of time. $\endgroup$
    – DanyloL
    May 7, 2020 at 13:59
  • $\begingroup$ Maybe my understanding is affected by a calculus approach: many little "energy pouring"s for some amount of time give us an overall energy poured. $\endgroup$
    – DanyloL
    May 7, 2020 at 14:01

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