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It's always been said that the work required to lift a weight up to some defined distance is the same wether you take 1 second or 1 minute to do it. What changes is the power. But, in order to do the process of lifting the weight in less time than 1 minute, we ought to increase the acceleration. If F=ma, increasing acceleration implies increasing the force. If W=Fd, increasing the force implies increasing the work. So, in order to do the process of lifting more rapidly, we must to increase the work. Therefore, if P=W/t, we cannot decrease the time without increasing the work. Is this right?

MIT 8.01 Physics I Classical Mechanics, Fall 1999
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Yes. You are right. if u need to increase P, then the ratio of w/t must increase-> which means if we need to increase w to reduce t. w is inversely proportional to t.
No, because, once accelerated, it will also decelerate, and then the force will be less. So work is the same.
I didn't catch your point, Vincent. Why it will decelerate? If I keep the force on the object constant during the entire trajectory, the acceleration will be constant, I suppose.

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GlornadoPM, to take the object up faster you don't increase the acceleration, you increase the speed!
Indraneel, thanks for the answer. But, anyway, the work-energy theorem says that the work is directly proportional to the square of velocity. So, since the mass is constant, more speed means more work. Right?
  • hba
@vincent-lyon.fr
Indraneel, the same force acting during the same distance would not produce a different final speed. If you argue that you must to increase the force, you are increasing the acceleration too. what I mean is: you apparently must to put more energy on the object in order to increase the power or, in other words, you cannot decrease the time of the process keeping the energy (work) constant.
  • hba
I am really sorry,I did not sign up for this course.
no problem, hba. Thank you!
  • hba
Welcome.
I can intuitively glimpse that a bigger force acting during less time would supply the same amount of energy to the object than a smaller force during more time. But I don't see it gazing the equations, since the work is just a function of the intensity of the force and the distance, not having a time component.
\[W=F.d\] \[F.d=ma.d\] \[d=\frac{ 1 }{ 2 }at ^{2}\] since initially at rest \[W=\frac{ 1 }{ 2 }m \left( at \right)^{2}\] so you see time appears in the work equation..you dont see it since it involves with acceleration to give an idea of the velocity
You have to consider the final kinetic energy too! If we are assuming a constant force, then yes, greater power means more work done ONLY if that force is applied through the whole distance. BUT, if your goal is to achieve a certain height, then you could use that same high power for LESS time, because the added kinetic energy will keep your object moving up, even when the force is no longer applied (even while it is decelerating.) This I believe is what Vincent was referring to.
Remeber: work is defined as the integral of F.dx (in 1 dimention), let's say, from point A to point B; if the force is conservative, the energy dispensed will be constant. When you lift up an object with constant velocity, the net force is 0!! being the velocity small or huge. And the 'net' work is 0 too. But, if you say some object is being lifited with a force greater than the weight of the object (mg), there will be an acceleration. In this case, you better look to work as energy. And if energy is to be conserved, it will be the same in both cases (object being lifted with constant or not-constant velocity). Whatever is the case, gravity will do a work measured by the formula W=m.g.h And certainly you are apliying a force to pull the object. If there's acceleration, then you just added energy to the system, which means, at the end of the AB course, there wil be extra energy, just it. The work that is the same is the one you use to lift something from A to B only, which is 'cancelled' by gravity.
u r first applying force but later to stop the object u have to decelerate too..... what vincent said is correct

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