Work and Gravitational PE

[ahyaa]

Hello,

 

I was wondering how signs are assigned to gravitational potential energy. Ug is positive when an object starts at rest and falls straight down. This makes sense to me because the gravitational pull of the earth (Fg = mg) pulls the object a distance of h (thus = mgh). This fits the work equation = Fdcos(theta). But if this is case, shouldn't Ug be negative, because Fdcos(theta) = mg*(-h)cos(theta), where -h is from the object moving along an arbitrary negative axis, and cos(theta) is 1 because the force and displacement vectors are parallel. Any enlightenment is appreciated.

[swansont]

The sign is arbitrary. If the statement of the problem does not restrict your choice, then you can choose your own coordinate system and convention for the PE. What's important will be changes in PE, and that will be unaffected by the sign.

[ahyaa]

Let's say we use an axis with positive pointing up.

 

Let's represent the stored energy due to its position (grav. potential energy) re-written as work (from work-energy theorem = delta E = Work).

 

If the object moves from h1 to h2, where h1 > h2 (that is the object is falling in this case) then the stored energy = F(parallel) * delta h * cos(0) = F(parallel) * (h2-h1) = -mg * -h (because h1>h2).

 

We thus have grav. potential energy = mgh (h = h2-h1) for an object falling from h1 to h2. I guess my question is, why is mgh positive for a falling object, shouldn't the object be losing potential energy, therefore mgh should be negative?

[swansont]

Let's say we use an axis with positive pointing up.

 

Let's represent the stored energy due to its position (grav. potential energy) re-written as work (from work-energy theorem = delta E = Work).

 

If the object moves from h1 to h2, where h1 > h2 (that is the object is falling in this case) then the stored energy = F(parallel) * delta h * cos(0) = F(parallel) * (h2-h1) = -mg * -h (because h1>h2).

 

We thus have grav. potential energy = mgh (h = h2-h1) for an object falling from h1 to h2. I guess my question is, why is mgh positive for a falling object, shouldn't the object be losing potential energy, therefore mgh should be negative?

You will not have -mg*-h. The PE is mgh. You have an extra minus sign.

 

The PE is not the work done by gravity, it's the work done to overcome gravity, i.e. The opposite sign

[ahyaa]

Ah, I figured it out. My main problem was 1) improperly using the dot product, which for vector(A) dot vector(B) = ||A||*||B|| cos(theta). Instead I used the vector for F as -mg and there should be no minus sign as you said. Extending this to the more general form of work W = Fd --> F and d are always positive. 2) The F in work is the external (not internal) force. As you said it's the force (and work from this force) to overcome gravity not due to gravity. Thanks!