Why is the potential energy for a body raised up by a height negative?

I tried deriving the potential energy of a body when raised at a height $h$ above the earths surface, using the formula:

$$PE = -W_\text{conservative}$$

where $W_\text{conservative}$ is the work done by conservative forces. I'm getting $-mgh$ for a body raised a height $h$ above the earths surface. Why is this negative?

• Could you maybe include your method for deriving that formula? There are many places you could have accidentally gained or lost a - sign so its quite hard to tell what the problem is without seeing your maths. Dec 14, 2014 at 13:22
• I am guessing he probably did this and arrived at the -mgh term. F=-mg downwards and dx= +h and F.dx = -mgh and probably forgot about the -ve sign in the equation of PE. anyway its not potential energy but "change in potential energy" isnt that right ? Dec 14, 2014 at 13:42

2 Answers

Remember the definition of work: $\vec F.d\vec x=Fx\cos\theta$. In your case $\theta=180^o$, thus $\cos\theta=-1$. That is the minus sign you were missing.

In your formula you don't calculate the potential energy, but the difference between the potential energy of the body at the lower level $P_E(h_0)$, and that at the higher level $P_E(h_1)$. Indeed, since $P_E(h_0) < P_E(h_1)$ you get $P_E(h_0) - P_E(h_1) < 0$ as expected.

Phenomenologically, starting from $h_0$, you have to invest some work to raise the object at the level $h_1$, s.t. $$P_E(h_0) + Work = P_E(h_1),$$ s.t. $$P_E(h_0) - P_E(h_1) = -Work.$$ Your mistake is that not the field forces do here the work for raising the body, but you do the work. So, for raising the object, you do work for adding to it potential energy.