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Say you have a 1kg mass at end of string and you raise the mass 1m at a constant speed. Then you lower the mass 1m. The amount of work to raise it is F*d = ma*1m = mg*1m = 1kg(9.8m/s/s)(1m) = 9.8J

But to lower it it is the exact same calculation but it doesnt make intuitive sense to me. At the start, I am holding the string that suspends the mass, and when I raise it I have to expend energy to raise it 1m But when I lower it back down I expend less energy. Its confusing me. thanks for any help

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When you raise the mass you do work, placing energy into gravitational potential this is your 9.8J.
When you drop the mass gravity does work, creating kinetic energy. If you drop the block 9.8J of work done will be done by gravity.
However as you are not dropping the block, but are lowering it you are resisting some of the acceleration due to gravity, and then later applying a counter acceleration greater than gravity to stop the block moving.
The work done by both you and gravity combined on the block is the force that accelerates and then decelerates the block to its new position one meter lower. Say it is 0.8J.
However considering just your work; you are working against gravity to reduce the acceleration to this level. You may need to put in something like 9J to achieve this.

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    $\begingroup$ No accelerations here, idealized constant v. The work up and the work down on the mass are identical in magnitude, opposite in sign. $\endgroup$ – bpedit Jan 26 '17 at 18:05
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You are correct that lowering it requires less energy than raising the mass. In fact lowering it requires no energy expenditure (on you part) at all (you could just turn it loose and let it drop). In falling, the mass can do work (useful or not) that is equivalent to the work you did in raising it. That is a property of all conservative fields including gravity.

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  • $\begingroup$ so you mean it is perspective.. it must be looked at as mass doing work, not me doing work..is that correct? $\endgroup$ – Mocumentals Jan 26 '17 at 16:12
  • $\begingroup$ When you raise it you are doing the work. When it falls (and hits something) it is doing the work. When you lower it so that it doesn't fall you doing some work, but not as much as when you raised it. $\endgroup$ – Lewis Miller Jan 26 '17 at 21:02
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First thing you need to know is that gravitational force is a conservative force and the net work done by conservative forces in a closed loop is always zero....so if you are taking it up and then down to initial state then the net work should be zero....from there you get that work done should be same but of opposite sign

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The magnitude of the work lifting the mass 1 m is exactly the same as that in lowering it 1 m. In both cases the force you exert to either lift or lower the mass at constant velocity is simply the wieght of the object, a constant. The distance is also the same in both cases, therefore the magnitude of Fd is identical.

What's different is the sign of the work. On the way up, F and d are in the same direction so the work is positive. The result of this is a gain in the mass's gravitational energy. When the mass is lowered, the force you are exerting is upward but the direction the mass is moving is downward. The product of these is now negative, the result is a decrease in the gravitational energy of the mass.

You are not exerting the only force on the mass, the pull of Earth's gravity is also acting. In this case, as you raise the mass the work done by the Earth on the mass is negative. It is equal and opposite to that of you lifting the mass making the net work done on the mass zero. This implies, according to the Work-Energy Theorem, there is no change in kinetic energy. Likewise on the way down.

As an aside; it doesn't matter that the mass is lifted at constant velocity as long as it is at rest (or the same velocity) at both ends of the problem. Any amount of greater work done in accelerating the mass at the start will be compensated by the lesser work in decelerating at the end.

In the case of lifting, you have done positive work on the mass-Earth system thereby increasing its gravational energy. In the case of lowering the mass, you perform negative work on the system lowering its gravational energy. In both cases the work done by the Earth on the mass is within the system and doesn't affect the gain or loss of gravitational energy of the system. In fact, within the system, any work done by the Earth on the mass is canceled by work done by the mass on the Earth.

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Further to the concept of negative work:

Human muscles are designed to do work; they are not designed to benefit from having work done on them. (Instead, we eat). This hides the fact that when the load is lowered, it is doing work on the thing lowering it.

For example, an electric motor uses electrical energy and does work in lifting an elevator car to the top of a building. The same "motor" will exert a resisting force to lower the car. But the motor is now acting as a generator, having work done on it to produce electrical energy that flows out into the "source" of electrical energy.

At some hydro-electric generating sites, some of the electricity is used to pump water up into an elevated storage pond. Later, when demand picks up, this water flows down through the same pump/turbines to do work in the generators to make electrical energy. (Google "pumped storage")

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