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...