Confusion regarding work Suppose that two forces with corresponding value are acting on the same object with mass m. Force A travels through a distance of 1 m, whereas the other a distance of 3 m. The work done by force B is 3 times than that of Force A. My question is why ? If the object after travelling through a distance of 1 m is already going, say 10m/s then the time required to push the object over another meter should be less than the previous one ( less time to apply the force means less momentum transferred which amounts to lesser velocity gains ) . Therefore the  magnitude of the final velocity (after 3 m) should not equal 3 times the starting velocity. This means that the kinetic energy when 3 meter has been reached shouldn't be 3 times the kinetic energy gained by the object after one meter. Where did I go wrong here?
 A: Kinetic energy is not proportional to the velocity $v$, but to $v^2$. If I understood you correctly, this could be what you overlooked. However, let's do the math thoroughly to see clearly what is going on:
At first, $m$ is at rest or moving at constant speed, in which case we transfer to the inertial frame of reference of $m$, where it is at rest. Now, starting at time $t=0$, the force $F$ acts on $m$ while $m$ covers $d_1 = 1 m$, afterwards the kinetic energy of $m$ is
$$
E_1 = F d_1~.
$$
While $F$ accelerates $m$, the distance covered is
$$
s(t) = \frac 12 \frac Fm t^2 \quad \Leftrightarrow \quad t = \sqrt{\frac{2ms}{F}}~,
$$
so if $s(t_1) = d_1$, then $t_1 = \sqrt{2md_1/F}$. At $t = t_1$, the kinetic energy is
$$
E_1 = \frac 12 m v^2(t_1) = \frac 12 m \left(\frac Fm t_1\right)^2 = F \cdot s(t_1) = Fd_1~,
$$
consistent with the previous result.
In the next step, from $t = t_1$ to $t = t_2$, $m$ covers a distance $d_2 = 3m$ and $F$ continues to act on it. The additional kinetic energy is
$$
E_2 = F d_2~,
$$
so the total kinetic energy at $t = t_2$ is
$$
E_{\text{tot}} = E_1 + E_2 = F(d_1 + d_2)~.
$$
Again, if $s(t_2) = d_1 + d_2$, then $t_2 = \sqrt{2m(d_1 + d_2)/F}$ and at $t = t_2$, the kinetic energy is
$$
E_{\text{tot}} = \frac 12 m \left(\frac Fm t_2 \right)^2 = F s(t_2) = F(d_1 + d_2)~,
$$
which is likewise consistent with the previous result.
Using all of this, let's sum everything up:

*

*As you pointed out, less time is required to cover a meter between $t_1$ and $t_2$. The average velocity between $t = 0$ and $t = t_1$ is
$$
\frac{d_1}{t_1} = d_1 \cdot \sqrt{\frac{F}{2md_1}} = \sqrt{\frac{d_1F}{2m}}~,
$$
while it holds
$$
0 < t_2 - t_1 = \sqrt{\frac{2m(d_1 + d_2)}{F}} - \sqrt{2md_1}{F} < \sqrt{\frac{2md_2}{F}}~,
$$
so between $t = t_1$ and $t = t_2$ the mean velocity is
$$
\frac{d_2}{t_2 - t_1} > \sqrt{\frac{d_2 F}{2m}} > \sqrt{\frac{d_1 F}{2m}} = \frac{d_1}{t_1}~.
$$

*Again as you pointed out, the ratio of the velocity at $t_2$ to the one at $t_1$ is
$$
\frac{\frac F m t_2}{\frac F m t_1} = \frac{t_2}{t_1} = \frac{\sqrt{2m (d_1 + d_2)/F}}{\sqrt{2md_1/F}} = \sqrt{1 + \frac{d_2}{d_1}} = \sqrt{1 + 3} = 2 \neq 3~.
$$

*However, the final kinetic energy $E_{\text{tot}} = F \cdot 3m$ is three times $E_1 = F \cdot 1m$. As I said initially, this is because the energy depends on the square of the velocity.

A: Maybe you do a real calculation with a mass of 1kg and force 2N which means accelleration $2\frac{m}{s^2}$ to the velocity $v=2\frac{m}{s}$ in $0.5s$
now you take a force of 6 N, an accellertion of $6\frac{m}{s^2}$  and you reach 3 m in 0,72s with the velocity $v=(2+6*0.72)\frac{m}{s}=6.32\frac{m}{s}$
3 times the force does not give you 3 times the velocity, you can calculate withe the equations of motion or with conservation of energy , both gives the same result.
A: There are some fundamental errors that need to be addressed in your question that I believe are causing you some confusion.

*

*You should not consider forces as being in motion.  A force may be applied to an object resulting in the motion of the object, but the concept of force itself is not something that moves through space with time.  Therefore when you say a force is moving through a distance of 1m or 3m you are creating a confusing situation.


*You should specify the direction of the forces. One can imply that both of these forces are oriented in the same direction, but this is never clearly stated.


*One should be careful in thinking about transferred momentum.  In classical mechanics energy can be transferred and momentum is always conserved, but thinking about transferred momentum can create some problems for you.


*In classical mechanics, kinetic energy is proportional to velocity squared (not velocity)


*You imply that since the object has greater velocity after moving one meter (when the first force falls off)  that the time to move it the next meter should be less and therefore less force is needed.  You are correct that the time interval will be less, but it will still move one meter under the influence of a constant force (in less time) and therefore the work done (from the second force) is exactly the same in moving the second (and third) meter as it was for the first meter.


*In the last section where you say "kinetic energy" gained it might be easier to think of this as the "work done".  When looking at it this way you should not expect the work done over 3 meters to be 3 times the work done over the first meter.  Why?  Because there were two forces in place during the first meter and only one force present for the final two meters of travel.


*If you are only considering the second force, then yes, the work done over the three meters (by the second force) is exactly three times the work done over the first meter.  The time it takes to move through each interval does not matter.  It is the distance moved under the influence of a force that determines the work done.
Hope this is helpful.  If I misunderstood your question I apologize, but - in the future - try and better think through the wording of your question.
