Recent content by Matt951

Thank you! I finally got the answer. I just keep on over complicating my formulas which makes everything so much harder than it needs to be

Okay so here are my calculations ΔPE = -ΔKE (my2*g*hy2)-(m*g*Δh) = -.5mv² (initial KE is 0) than multiply over -2 from KE and divide by m (-2(my2*g*hy2)-(m*g*Δh))/m = v²

Ok so the left side's mass is (.44/1.3)*2.4 = .81 kg and its change in height is zero since the whole rope goes down by the length of the left height and the right side's mass is (.86/1.3)*2.4 = 1.59 kg and its change in height is just the length of the left side When you ask for final distance...

So you're saying I should do this: Uο=ΔU mghy1+mghy2=ΔU and since ΔU= -ΔKE mghy1+mghy2=-.5mv² get rid of Ms and take out gravity on the left g(hy1+hy2= -.5v² multiply -2 on both sides with the negatives of -2 and gravity canceling 2g(hy1+hy2)= v² then square root v=√2g(hy1+hy2) or should I keep...

U + Ke = Uο + Keο mgΔh + .5mVf² = mg(Δh = 0) + .5mV²(Vi should equal 0) cancel all the Ms since there in each individual formula gΔh = -.5Vf² multiply -2 on each side to get rid of the .5 on the right and then square root Vf = √-2(-9.81)*Δh so you're saying that my Δh should begin in the center...

Homework Statement A limp rope with a mass of 2.4 kg and a length of 1.3 m is hung, initially at rest, on a frictionless peg that has a negligible radius, as shown in the Figure. y1 is equal to 0.44 m. What is the vertical velocity of the rope just as the end slides off the peg? Homework...