# Homework Help: Gravitational Potential Energy Derivation question?

1. Sep 30, 2010

In deriving the gravitational potential energy term I have a question.

$$W \ = \ \int_{r_1}^{r_2} \overline{F}( \overline{r}) \cdot \,d \overline{r} \ = \ \int_{y_1}^{y_2}mg \,dy$$

$$W \ = \ \int_{y_1}^{y_2}mg\,dy$$

$$W \ = \ mgy_2 \ - \ mgy_1$$

[PLAIN]http://img696.imageshack.us/img696/2303/workvk.jpg [Broken]

$$W \ = \ mgy_2 \ - \ mgy_1 \ = \ U_{grav}_2 \ - \ U_{grav}_1$$

I think I understand that y2 < y1 and that is the
reason why people write the above as:

$$W \ = \ mgy_1 \ - \ mgy_2 \ = \ U_{grav}_1 \ - \ U_{grav}_2$$

but is it such a crime to just be aware of the y2 < y1 and
write the equation in the more logical fashion that the straight calculation gives
you. To me it seems similar to how you rewrite the equations of constant
acceleration the standard way the calculus shows them and you mentally
set g = - 9.8 m/s²

Just like to hear some thoughts on this, thanks!

Last edited by a moderator: May 5, 2017
2. Sep 30, 2010

### alphysicist

There is a problem here with the second integral. Remember that the dot product depends on the direction of the vectors, and that this force is downwards.

Last edited by a moderator: May 5, 2017
3. Sep 30, 2010

I don't see the problem, the force is w = mg and the direction is downwards so both
the force and direction are in the same direction ergo there should be no minus.
What am I missing?

$$\overline{F} \ \cdot \ d \overline{r} \ = \ mg \ \hat{j} \ \cdot \ (\ dx \ \hat{i} \ + \ dy \ \hat{j} \ + \ dz \ \hat{k} \ )$$

$$\overline{F} \ \cdot \ d \overline{r} \ = \ mg \ dy$$

?

4. Sep 30, 2010

### alphysicist

The dy is positive; the limits that you have put in there handle the fact that you are actually moving downwards. (At least, that is why you are getting a different answer. There are other ways to interpret it, but that is the easiest.)

(Also, you know the work done by gravity as the box moves down is positive, and yet you are getting a negative work.)

5. Sep 30, 2010

Hmm, I'm still a little confused I must admit. Thanks a lot for the help, I think if you could
help me as I explain a specific example I think it'll be easy to generalize from it.

Say I'm dropping a 10kg box from a point 10m above the ground, say y1 = 10m
to a point 2m above the ground, y2 = 2m. The force acting on the box is
$$\overline{F} \ = \ - \ m \ g$$

$$W \ = \ \int_{y_1}^{y_2} \ \overline{F} \ \cdot \ d \overline{r}$$

$$W \ = \ \int_{10}^{2} \ (- \ m \ g ) \hat{j} \ \cdot \ dy \ \hat{j}$$

$$W \ = \ - \ m \ g \ y \ |^{2}_{10}$$

$$W \ = \ - \ m\ g \ (\ 2 \ - \ 10 \ )$$

$$W \ = \ + \ 8 \ m\ g \$$

W = 800 J (g = 10m/s² for convenience).

I assume the calculation for the general term for potential energy follows a similar
pattern? I mean, I should count F as (- mg)i and then the minus will cancel out if
we are moving in the direction of motion or remain if we go against gravity?

If I do this the derivation goes as follows:

$$W \ = \ \int_{y_1}^{y_2} \ \overline{F} \ \cdot \ d \overline{r}$$

$$W \ = \ \int_{y_1}^{y_2} \ (- \ m \ g ) \hat{j} \ \cdot \ dy \ \hat{j}$$

$$W \ = \ \int_{y_1}^{y_2} \ - \ m \ g \ dy$$

$$W \ = \ mgy|^{y_1}_{y_2}$$

$$W \ = \ mg_1 \ - \ mgy_2$$

$$W \ = \ U_1 \ - \ U_2$$

$$W \ = \ - \ (U_2 \ - \ U_1)$$

and voila?

6. Sep 30, 2010

### alphysicist

Looks good to me!

7. Sep 30, 2010