Recent content by jvalencia

That is vastly more intuitive! Thank you. Any point in the rope moves a distance equal to the difference in the hypotenuses of triangles x1 and x2. Therefore ## \sqrt\frac {2F(A-B)} {m} = v ## Where A and B are the hypotenuses of the aforementioned triangles.

What are appropriate ways to thank people who help me on this site?

My understanding of calculus is pretty shaky; I'm only in my first semester. Would this be the way to set up that integral? $$ \int F \cos\theta dx $$ $$ \theta = \arctan\frac{2}{x_1 - x_2 - x} $$ Also, I don't really see what you are hinting at with that last bit. I know that the tension force...

Here it is (Mentor's note: edited post to expand thumbnail image)

Thank you for telling me that. I shall figure out how to do so shortly

Homework Statement The figure shows a cord attached to a 2 kg cart that starts from rest and can slide along a frictionless horizontal rail. The right end of the cord is pulled over a frictionless pulley at height h = 2.0 m above the point of attachment of the cord to the mass, so the cart...

Hi all, I'm Jackson. I have no particular talent for physics and a limited understanding of calculus, so AP Physics C is kicking my butt. I'm hoping I will be able to learn on this forum and perhaps even help others along the way once I figure out the treacherous path that is physics.