Question

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?

Answer 1

Answer: $\Delta E_{pot} = -mgh$

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

Answer 2

"-mgh" would be the change in potential energy. A further explanation is provided below.

As we know,

The change in the potential energy is obtained follows:

→ $\Delta U = -\frac{mgh}{3}$

Now,

→        $= -\frac{3mgh}{3}$

→        $= -mgh$

Thus the above response is appropriate.

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