Work done in the presence of gravity

Question

Consider a system where a block is to be raised from the ground to a height of 1 meter above the ground. It is given that when the force acting on the block and its displacement act in the same sense, the block gains energy. On the other hand, when both quantities act in opposite senses, the block loses energy.

Back to our example. At the ground reference level, let's assume the energy of the block is zero (no potential energy and no kinetic energy). As the block rises towards the 1-m height, it is acted upon by two equal and opposite forces: the lifting force and weight. The lifting force and the displacement are in the same sense so energy is gained by the block. However, weight and displacement act in opposite senses so the block loses an amount of energy equal to that supplied by the lifting force. Now, the block is at 1-m height with zero net gain in energy.

The question is: The gravitational potential energy of the block clearly increased since there was a change in elevation. However, we assumed in the previous paragraph that the block did not gain or lose any energy. So where is this energy coming from?

Answer 1

You seem to have a misconception. Work done is not the change in total energy. Work done is the change in kinetic energy only.

$$W=\Delta E_k$$

So, as you point out, since they're equal and opposite forces, no kinetic energy is gained. In fact, the block was at rest at the beginning and it is at rest when lifted.

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It is a common misconception year by year. Usually badly explained. But the work-energy theorem states that work equals change in KE, only KE.

Then what? Well, consider splitting the work term in two parts: work done by conservative forces and work done by non-conservative forces, i.e. $W=W_C+W_{NC}$

Only the conservative forces allow writing $W_C=-\Delta E_p$, and the sign matters.

Gravity is conservative. Your force is not, even if you're compensating it temporarily.

Consequently, you have

$$\Delta E_k = W_C + W_{NC} = -\Delta E_p + W_{NC}$$

Or, if you move all energies to the left:

$$\Delta E_k + \Delta E_p = W_{NC}$$

That's mechanical energy:

$$\Delta E_m=W_{NC}$$

If there are no non-conservative forces, $\Delta E_m=0$ and energy is conserved, hence the name.

When energy is not conserved, it is because there's some non-conservative force changing it. Especially friction, that's teh most frequent.

Answer 2

In order to start the block moving from rest it is necessary to exert, at minimum, an upward force greater than the downward force of gravity. Once you get the block moving the upward force can be reduced to equal the downward force of gravity. Although the net force is now zero, the block continues upward at constant velocity (and constant kinetic energy).The upward force is in the same direction as the displacement of the block so work on the block is positive, and the block gains gravitational potential energy.

Just prior to reaching the 1 meter height, you reduce the upward force to slightly less than the downward force of gravity. The block now decelerates. The net force is downward but the displacement continues upward. Therefore the work being done on the block is negative (it loses kinetic energy) but its potential energy continues to increase as long as the motion is upward. If the block velocity is zero at 1 meter, the loss of kinetic energy at the end equals the gain at the beginning so that the total change in kinetic energy when it reaches 1 meter is zero. The block has gained only gravitational potential energy.

Now you can do one of two things.

You can release the block allowing it to free fall. The downward force of gravity is in the same direction as the displacement so gravity is doing positive work on the block giving it kinetic energy. However, at the same time, the block is losing an equal amount of gravitational potential energy. When it reaches the ground all the potential energy is now kinetic energy.

Instead of releasing the block you can also lower it by reducing your upward force to slightly less than the gravitational downward force. Once the block starts moving downward you can increase the upward force to equal the gravitational force so the block moves downward at constant velocity. The force you exert is now upward but the displacement of the block is downward, so the work done on the block is negative (the block is losing gravitational potential energy).

Before the block reaches the ground you can increase the upward force to slightly more than the downward force of gravity causing the block to decelerate so that its velocity is zero when it reaches the ground, at which point it possesses neither gravitational potential energy or kinetic energy.

In the first scenario (dropping the block) there is merely a conversion of gravitational potential energy to kinetic energy when it reaches the ground. That kinetic energy is the result of gravity doing work on the block. But the original potential energy was due to you doing work on the block.

In the second scenario (lowering the block), work is done by the block on you and it loses gravitational potential energy.

Hope this helps.

Answer 3

The work done by gravity is thought of as change in potential energy because, unlike the lifting force, the gravitational force is conservative. I recommend you to look at these links:

• Similar question in different forum

• Conservative force page on Wikipedia

Answer 4

This problem is more likely to deal with the way questions would be posed to you, I guess. Because I do not think you understood work done incorrectly. Except that, work done is only dealing with a change in energy.

Consider a block at rest on the floor (with potential energy being zero - this is a matter of convenience, since we will only measure a change in energy at the end). When we raise the block and keep it at rest at some height $h$ above the ground, and it remains at rest, it will not have any potential energy - because I have raised it (counteracting gravity).

If the body had potential energy, it would have immediately transformed into kinetic energy and would have been falling. This happens only when I released it from rest.