Recent content by KvnBushi

Your first equation for Kinetic Energy is incorrect. The Kinetic Energy of a rolling object is \frac{1}{2} M w^2 + frac{1}{2} Mv^2 where 'w' is rotational speed and 'v' is translational speed. I would take another look at the book until it makes more sense to you.

Homework Statement A war-wolf, or trebuchet, is a device used during the Middle Ages to throw rocks at castles and now sometimes used to fling pumpkins and pianos. A simple trebuchet is shown in Figure P8.77. Model it as a stiff rod of negligible mass 3.00 m long and joining particles of mass...

It is actually spelled: Kropotkin Peter Kropotkin Errico Malatesta Mikhail Bakunin I would provide links through WIkipedia except for my slow connection. Here is a good online anarchist library: http://dwardmac.pitzer.edu/Anarchist_archives/index.html

I am certainly not going to waste my time defending theories and practices that have endured a century's worth of logical debate and thought, but before you spout off that Anarcho-Syndicalism was the result of "young pot heads", which is HIGHLY offensive to historical evidence, please research...

This had the potential to be a pretty interesting thread.

I am going to try this way as well when I get back from eating. Cheers!

SOLVED W(x) = \int 8e^{-2x} = -4e^{-2x} + C ( I forgot the C earlier) W(x) = U_i(x) - U_f(x) -4e^{-2x} + C = 5 - U_f(x) U_f(x) = 5 - 4e^{-2x} - C SOLVE FOR C U(0) = 5 = 5 - 4(1) - C C = 4 SOLVE FOR U(x) U_f(x) = 5 - 4e^{-2x} - 4 U_f(x) = 1 - 4e^{-2x}

[SOLVED] Potential Energy as a function of x Homework Statement Take U = 5 at x = 0 and calculate potential energy as a function of x, corresponding to the force: 8e^{-2x} Homework Equations W_{net} = U_i - U_f W = \int_a^b F_x dx The Attempt at a Solution \int 8e^{-2x} dx...

Isn't the distance 6m ?

I am not following this very clearly. Would you do something like this: \frac{80kg}{10m} = 8 \frac{kg}{m} and then: W = \int 8y dy 9.8 J AAAAHHHHHHHHH! THIS FORMATTING IS MAKING ME LOSE MY MIND! IT IS RUINING MY HOUSEHOLD! I'm not understanding how you would do this one...

Heheheh, point taken :) . Katchum, it was difficult for me to understand your alternate method. If you aren't worried about it, neither am I :cool: . If you want to figure it out, I'm willing to help out as well but I'd need you to repost with more literal equations. Take care! :approve:

It is not the cross product, it is the dot product: W = \Delta \vec r \cdot \vec F

Awesome! THANK YOU very much, nrqed. If you would like help with drums, fighting, DHTML or elementary linux, I'd be happy to return the favour.------------Do you understand it as well, katchum?

Thank you for the information, nrqed. I tried this in my 3rd attempt at problem (c): \int_0^5 x dx = 1/2 x^2 |_0^5 = 1/2 (5)^2 = 12.5 J correct answer: 66.7 J

[SOLVED] diff. paths of a Force in xy-plane Homework Statement A force acting on a particle in the xy-plane is given by \vec{F} = (2yi + x^2j) where x and y are in meters. The particle moves from the origin to a final position having coordinates x = 5.00m and y = 5.00m...