Recent content by AionTelos

For anyone else wondering here was the solution I came too. KE_i = PE_f + KE_f \frac{1}{2}mv_i^2 = mgh + \frac{1}{2}mv_f^2 h = -2R v_i^2-4gR = v_f^2 N+mg=\frac{mv^2}{R} N=\frac{m(v_i^2+4gR)}{R}-mg m=1.8kg v_i=11\frac{m}{s} g=9.8\frac{m}{s^2} R=2m N=\frac{1.8((11)^2-4(9.8)(2))}{(2)}-(1.8)(9.8)...

I think I got it, I'm using the velocity at the bottom of the loop instead of at the top, will post the correct solution once I get it.

Homework Statement A block of mass m = 1.8 kg starts at rest on a rough inclined plane a height H = 8m above the ground. It slides down the plane, across a frictionless horizontal floor, and then around a frictionless loop the loop of radius R = 2.0 m. On the floor the speed of the block is...