# Variance between the energy required to lift/push a particular object to any height, and the potential energy of that object at that particular height

As shown in figure we have two objects , 1 and 2 . Each have a mass of 10 kg and height 1 meter. Suppose i lift object 1 to a height 1 meter. The energy required to lift object 1 is E= mgh ( m is mass, g is gravity , h is height ) E= 10×9.81×1 So E= 98.1 j. Now potential energy of this object is PE= mgh , = 10×9.81×1 so PE = 98.1 j. This shows that the energy required to lift the object upto any height is equal to the potential energy of that object at that height.

But , depending on height of the object and our consideration , there may be a difference between , the energy required to lift the object and the potential energy gain of that object. To understand this , now we talk about object 2. Object 2 has 10 kg mass and 1 meter height. Now suppose and consider that object 2 has two parts , A and B as shown in figure , mass of A is 5 kg and height 0.5 meter , mass of B is 5 kg and height is 0.5 meter. So together mass is 10 kg and height is 1 meter.

Now i lift object 2 upto 1 meter . As the mass of both objects ( 1 and 2 ) is same , the energy required to lift object 2 is also 98.1 j. Now we calculate the potential energy of object 2 considering the two parts A and B.

PE of B = mgh= 5×9.81×1= 49.05 j

PE of A ,

part A is on and above part B , the height of part B is 0.5 meter , so the total height for part A is 1+0.5 = 1.5 meter.

So PE of A= mgh= 5×9.81×1.5= 73.575 j.

So the total potential energy of object 2 = PE of A+PE of B = 122.625 j.

So from above calculations.

Energy required to lift object 1 = 98.1 j

Potential energy of object 1 = 98.1 j

Energy required to lift object 2 = 98.1 j

BUT

Potential energy of object 2 = 122.625 j

So in case of object 2 , the potential energy gain is more than the energy which was required to lift this object. It means that if any object is at any height , and if we assume and consider that object in parts , then it's potential energy is always more than the energy which was required to lift that object to that particular height.

So here we can see the violation of energy conservation.

Does it means that , by this way the law of conservation of energy is violated ?

In this problem, calculating the change in potential energy is very straight forward as it is just:

$$U_g = m_0 * \Delta h * g$$

To solve for the total GPE, you could consider the top of the block as having all of the PE. From there, you would consider its final height to be 2m at solve with the same equation. That would be incorrect, however, as (as far as I am aware) you find GPE at the center of mass. This is shown with:

$$U_g = \sum_{i} m_i g k_i$$

Which can be simplified to:

$$U_g = m * g * k_c$$

So, in this situation, the law of conservation of energy is not violated. Or, in any situation for that matter.

For both blocks, the final GPE would be:

\begin{align} U_g &= 1.5 * 9.8 * 10\\ &= 147 J \end{align}

It would be more appropriate to set the height as 1 meter, however, as both block a and b can fall no more than one meter.

If there's anything wrong with what I wrote, please don't hesitate to let me know.

• I said that because i have compared two things , 1) work done and 2) result of work done. In your answer you have just calculate the result of work done ( final GPE for both blocks 147 j ) , but not calculate the work done. So please calculate the energy required to do the work and compare it with final GPE. When I calculate , assuming the position as initial position , it was 98.1 j . Even after adding the energy required to lift and put A over B ( 24.525 j ) , it is 122.625 j . So we see here , final GPE is more than energy required to lift that object. Commented Oct 31, 2020 at 9:39

Your mistake is figuring the energy required to lift A 1 meter and then stating its PE as 1.5 meters. In your diagram A can only fall 1 meter to be back at its original position on top of B. If you want it to fall the full 1.5 meters to land beside B then you will need to add the energy originally needed to lift A on top of B. Then you will find energy to lift will equal PE.

I think hard about it and see that this example could show the difference between the energies. Please see the below picture.

As shown in the picture , there is a pipe. Height is 1 feet or 12 inches Diameter is 6 inches It has a plunger at the bottom and a bowl attached to the top corner. Bowl is 8 inches above the ground. Bowl diameter is 6 inches and height is 4 inches. Pipe is fully filled with water. As per the dimensions ( 6" diameter and 12" height ) of the pipe , the amount of water in pipe is 5.56 litre.

Now suppose i lift the water ( mass ) up by pushing the plunger in upward direction. When i push the plunger in upward direction , the water overflows from the top open end of pipe and goes to the attached bowl.

Suppose i pushed the plunger upto 4 inches .

The energy ( E ) required to lift the water (mass) upto 4 inches is E = mgh , putting the values , ( 4 inches = 0.1016 meter ) Consider 1 litre = 1 kg E = 5.56×9.81×0.1016 E = 5.54 joules

Plunger is pushed upto 4 inches , so as per dimensions ( 6" diameter and 4" height ) the amount of water overflow ( Displaced ) is 1.85 litre. Now , the remaining water in the pipe is 3.71 litre. Here we see that , when the mass ( water ) is lifted , it splits into two parts. Now we calculate the potential energy ( PE ) of the water ( mass )

1. Water in the pipe is 4 inches above the ground so , height is 4 inches = 0.1016 meter PE of water in pipe , PE = mgh , putting values PE = 3.71×9.81×0.1016 PE = 3.6977 joules

2. Water in the bowl is 8 inches above the ground so , height is 8 inches = 0.2032 meter PE of water in bowl , PE = mgh , putting values PE = 1.85×9.81×0.2032 PE = 3.6877 joules

So , total potential energy is PE = 7.3854 joules

So from above calculations we see that the potential energy ( PE ) of the mass is more than the energy which was required to lift that mass.

So , we may say that when any mass is lifted , and if it splits into parts and those parts occupy different positions then , the potential energy gain of such mass is more than the actual energy which was required to lift that mass.