Question on work

[hotcommodity]

If work is definied as:

 

[math]W_F = \int \vec{F} \cdot d\vec{r} [/math]

 

And the dot product is:

 

[math] |F||dr| cos(\theta) [/math]

 

then I have [math]W_F = \int \vec{F} \cdot d\vec{r} = \int |F||dr| cos\theta[/math]

 

Let's say that I wanted to show that the work done by gravity on a falling object, from [math] h_0 [/math] to [math] h_f [/math] is [math] W_{grav} = - \Delta U [/math] by using the integral of the dot product above.

 

The displacement vector is in the same direction as the force, so [math] cos\theta = 1 [/math]. Now I have the integral of the magnitude of F, times the magnitude of dr. How do you end up with a negative in front of the change in potential energy when you have two magnitudes being multiplied? Any help is appreciated.

[swansont]

[math]W_{grav} = - \Delta U [/math]

 

Is this representing the work done by gravity, or the work done against gravity?

[hotcommodity]

That would be the work done by gravity. The work done by any conservative force is always equal to the negative of the objects change in potential energy, right?

[swansont]

Right, that's how we use potential energy. It's related to how much work someone would have to do to move the object around, and the applied force would be in the opposite direction of the conservative force, so you get a negative sign.

[hotcommodity]

But as far as the integral goes, I'm confused where the negative comes in? Is that an inappropriate way of finding the work done?..

[swansont]

Ah, OK. The negative sign comes from the way that PE is defined, not from that integral. The work-energy theorem states that [math]W = \Delta KE[/math] and conservation of (mechanical) energy dictates that if [math]W_{noncons} = 0, \Delta KE + \Delta PE = 0[/math] which leaves you with [math]W_{cons} = -\Delta PE[/math]

[hotcommodity]

I see. But there's no way to use that definition of the dot product to find that its negative change in potential energy? I was told that some work definitions are for special cases.

 

Edit: btw, I missed an integral in the opening post, sorry.

[tvp45]

I see. But there's no way to use that definition of the dot product to find that its negative change in potential energy? I was told that some work definitions are for special cases.

 

Edit: btw, I missed an integral in the opening post, sorry.

 

What are your integration limits?

[hotcommodity]

I'd be integrating from the initial height to the final height.

[swansont]

I'd be integrating from the initial height to the final height.

 

That will give you a - sign if you are using the usual coordinate system with the y axis being up.

[hotcommodity]

I know there are a few equations used to find the work done on an object, I'm just trying to use all of them to prove the same thing. But I'm still missing the negative sign when I use the magnitude definition of the dot product.

 

Using vectors I'd have:

 

[math]W_{grav} = \int^{h_f}_{h_0} \vec{F} \cdot d\vec{r} = \int^{h_f}_{h_0} <0, -mg, 0> \cdot <dx, dy, dz> = -mg \int^{h_f}_{h_0} dy[/math]

 

Evaluate the integral:

 

[math] W_{grav} = -mg(h_f - h_0) = -\Delta U [/math]

 

Using components I'd have:

 

[math]W_{grav} = \int^{h_f}_{h_0} \vec{F} \cdot d\vec{r} = \int^{x_f}_{x_0} F_x dx + \int^{y_f}_{y_0} F_y dy + \int^{z_f}_{z_0} F_z dz [/math]

 

Zero force occurs in the x and z directions, so I get:

 

[math]W_{grav} = -mg \int^{h_f}_{h_0} dy = -mg(h_f - h_0) = -\Delta U [/math]

 

But when I use the definition of the dot product using magnitudes, I get:

 

[math]W_{grav} = \int^{h_f}_{h_0} \vec{F} \cdot d\vec{r} = \int^{h_f}_{h_0} |mg||dr| cos\theta = mg \int^{h_f}_{h_0} dr = mg(h_f - h_0) = \Delta U [/math]

 

The negative is missing in this case because I only use magnitudes. I'm not sure where my reasoning is flawed.