Potential energy change is not negative? [closed]

Question

$\Delta U = -(W_{earth} + W_{ball})$

$W_{ball}$ is almost 0, as earth's displacement by the falling ball is super small, so $\Delta y$ of the earth could be negligible and $W_{ball} = 0$. so:

$\Delta U = -(W_{earth})$

$U_{f} - U{i} = -(mg(y_f - y_i))$

When object falls down from 100m to 80m, $y_f - y_i$ is negative. The negative sign of this cancel the negative sign in the start of the equation, so we're left with:

now, type in 100 and 80, and we will get $20mg$ which seems change in potential energy appeared positive. It should be negative though as it decreased from 100m to 80m.

Answer 1

This question is not so silly, there is a misunderstanding. The work is a dot product between the force $\textbf{F}$ and the displacement $\textbf{s}$. The direction of both is crucial to avoid a problem of sign. The direction your $\vec{y}$ axis as it seems to be is originated from the ground to higher altitudes, as the object is falling : $\textbf{s}$ and $\textbf{F}$ have the same direction $W(\textbf{F})=-mg(y_f-y_i)=mg(y_i-y_f)>0$

A negative work of a falling object is absurd. Therefore $\Delta U <0$.