# If the change in potential enegry is equal to the negative of the work done, then this principle isn't consistent here in the case freely falling body

Let us assume that a body of mass $$m$$ falls from height $$h_1$$ to $$h_2$$ :

Here the Work done by gravitational force (Conservative force) is :

$$\mathrm{Force \ ×\ Displacement} = mg \ (h_2-h_1) \tag1$$

Also the change in potential energy is:

$$\mathrm{Final \ potential \ energy \ -\ Initial \ potential \ energy} = mgh_2 - mgh_1$$

$$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = mg(h_2-h_1) \tag2$$

Thus $$(1)=(2)$$, but according to the principle,

The change in potential energy should be equal to the work done by a conservative force.

(here the conservative force is gravitational force).

The problem is that you are not considering the sign of g, which is a vector. If $$h_2 < h_1$$ then you are saying that the positive direction of h is straight up. But notice that the acceleration g is a vector and points straight down, given the orientation: Straight up = positive and Straight down = Negative, then g is negative, because it points straight down, so the work done by the force is:

$$Work = m * (-g) (h_2 -h_1)= -mg(h_2 -h_1)$$

You can also think like this: The force of gravity is acting in the same direction of the movement, so the work has to be positive. Since $$h_2 - h_1$$ is negative, then you put the negative sign on g.

$$\Delta E_p = -Work = mg(h_2-h_1)$$

Which is the expected result

• So all I had to do was take the acceleration due to gravity as negative (-g) right? Commented Feb 15, 2022 at 19:08
• Yes, but that's only because your h-axis is positive straight up. If, for some reason, you decided to invert the axis and the positive direction was straight down, g would be positive, and h2-h1 would be positive because the height would be increasing straight down. Remember to define your coordinate system properly before solving problems and you're good. Commented Feb 15, 2022 at 19:39
• Thank you so much! Commented Feb 17, 2022 at 3:34

Taking altitude as z-axis, growing upwards, the force is negative because points downwards: $$F = -mg$$. So the work is positive: $$-mg(h_2-h_1) = mg(h_1-h_2)$$

And the variation of potential energy is negative: $$mg(h_2-h_1)$$