Recent content by gggorillaz

Yay! thank you so much for helping!

Ok sweet, so i would find integral of sqrt(3)(x^2/2 -2x) which = sqrt(3)/2(x^3/3 - x^2). now if i plug 2m in for x i find sqrt3 /2(8/3 - 4) which =-2sqrt(3)/3 right?

Ohhh right, I see now, it would now be mgh=1/2mv^2+mg(height of loop) so h=1/2(r)+h =1/2(20)+40 = 50m right?

Homework Statement A force is applied to a block downwards at 30 degrees above the horizon line. the Force = x^2 - 2x (Newtons). Find work over the first 2 meters Homework Equations Work = integral(F(x)dx) Work = integral(Fcos(theta)dx) Work = F(x)delta(x) The Attempt at a...

iIs it that i have a skewed understanding of what centripetal force equals? maybe that ma=mv^2/r but my a isn't = g in this case? Should i be getting h=2diameter? if so, how do i get there?

Homework Statement A mass of 100kg is h meters high on a track that extends into a loop that has a radius of 20m. Assume track is frictionless. I need to find minimum height for the mass to make it around the loop without falling off or going backwards. Homework Equations Conservation...