Physics: Considering that the potential energy of the block at the table is mgh/3 and that on the floor is −2mgh/3, what is the change in potential energy δu of the block if it is moved from the table to the floor?

Considering that the potential energy of the block

taythestargazer71

11.03.2022

$\Delta E_{pot} = -mgh$

Explanation:

The change in potential energy is the difference between the energy at the floor level and the energy at the level of the table:

$\Delta E_{pot} = E_{pot}(floor) - E_{pot}(table)\\\Delta E_{pot} = -mg\frac{2}{3}h - mg \frac{1}{3}h=-mgh$

Given these two points and an unknown m and h, the change in the potential energy due to the "fall" from table to the floor is -mgh

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