Why does less force correspond to less work done?

Question

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.

Answer 1

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.

Answer 2

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.