Jump to content


Senior Members
  • Posts

  • Joined

  • Last visited

Profile Information

  • Location
    Not Texas
  • College Major/Degree
    Student of Engineering Physics


  • Atom

Recent Profile Visitors

The recent visitors block is disabled and is not being shown to other users.

hotcommodity's Achievements


Atom (5/13)



  1. I'd pursue a degree in mathematics. It seems like the more math you know, the easier problem solving becomes.
  2. 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.
  3. I'd be integrating from the initial height to the final height.
  4. 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.
  5. 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?..
  6. 61% You got 13 of the 21 people correct, and you did better recognizing the virginity of guys. Overall, you guessed better than 64% of all test takers. Girls have a poker face when it comes to virginity >:|
  7. 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?
  8. 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.
  9. I'm not sure what you mean by "motion for the 2nd..." If you're talking about motion in general, you could do a presentation on the x-prize. http://www.xprize.org/
  10. You can't really expect any institution to publicly back what Dr. Watson is saying, no institution is interested in having protestors at their doors. Chances are he wouldn't have declared this theory if there was any scientific argument that could readily challenge his belief. If he's not given a chance to explain his theory any time soon, it will probably take longer to debunk.
  11. Thank you for the reply. Ok, I get that it won't reach an absolute maximum, but if I consider "x" to be the distance that the spring compresses due to the block, can I think of that point as an end point graphically? In other words, can I plug in y = -[(.75m + x) sin(30°)] for the velocity function to obtain the maximum speed?
  12. Thank you for the reply. If I define the downward slope of the incline to be the positive x-axis, then the block will have a positive velocity. I defined h=0 to be the point at which the block starts accelerating, and [math]h_f[/math] to be the point at which the block begins decelerating. So I'd have: [math]0.5*m*v^{2}_f + mgh_f= 0[/math] [math]0.5*m*v^{2}_f = -mgh_f[/math] Now v final is a function of [math]h_f[/math] [math]v^{2}_f = \sqrt{-2gh_f}[/math] [math]\frac{dv_f}{dy} = \frac{-g}{\sqrt{-2gh_f}}[/math] Is this correct?
  13. You have a point, and I suppose I'm a little naive when it comes to the importance of voting. But lets say I research all of the candidates, follow all of the debates, and subsequently vote. How many voters would give that same consideration to each candidate in an effort to make an informed decision? I think alot of Americans, if not most, think of voting as nothing more than an American tradition, like the 4th of July. They label themselves "Republican," "conservative," "Democrat," and "liberal," and they let these titles, rather than information, decide their vote. I'm not very confident in the opinion of the masses. And I can't help but think that my vote would effectively work to counterbalance the votes cast by those who simply don't care to make an informed decision.
  14. Thank you for the reply. I'm still having a bit of trouble with this problem. If I understand what you're saying, the block will achieve its maximum speed right before it begins decelerating (and graphically, this makes sense), and it begins decelerating as it hits the spring. For part "a" of the problem, I found that the block does not compress the spring, x is equal to zero. So I think I only have to consider the movement of the block from the time it begins accelerating to the part right before it hits the spring. I'm not sure how to find this maximum speed, because in the dv/dy equation above, I have "g" in the numerator, and that's just a constant, it can never equal zero....
  15. I have a question on a homework problem having to do with a block accelerating toward a spring down an inclined plane. The problem is this: A 1.0 kg block starts at rest and slides a distance of 0.75 m down a frictionless 30 degree incline (with respect to the horizontal) where it hits a spring whose spring constant is 9.8 N/m. (No need to comment on my windows paint skills). Part "a" asks me to find the distance the spring is compressed, and I figured that out by using the the fact that the total mechanical energy of the system is conserved. Part "b" is what I'm having trouble with. It asks "where does the block achieve maximum speed during this slide." One of the hints my professor gave was to find a function of velocity and find its maximum by taking the derivative of that function. So this is what I did: I used the fact that mechanical energy is conserved from the time the block begins accelerating to right before it touches the spring. The energy terms below only deal with the block. [math]E_0 = E_f[/math] [math]U_f + K_f = U_0 + K_0[/math] The final potential energy, and the initial kinetic energy are zero if I take the point of the spring to be height = 0. [math]K_f = U_0[/math] [math]0.5*m*v^{2}_f = mgh[/math] [math]v_f = \sqrt{2*g*y}[/math] [math]\frac{dv_f}{dy} = \frac{g}{\sqrt{2*g*y}}[/math] It appears that it would achieve its maximum speed when the acceleration of the block is zero. If that's the case, this would be the point where the block is stopped by the spring. But if it's stopped, how can it have a maximum speed. Am I doing this wrong or am I missing a concept? Any help is appreciated.
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.