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**thedoge** Imagine a pulley rigged with a separate mass on each side connected by a string.

m1 = 2.00 kg and m2 = 8.00 kg

The masses of the pulley and string are negligible by comparison. The pulley turns without friction and the string does not stretch. The lighter object is released with a sharp push that sets it into motion at vi = 2.80 m/s downward.

How far will m1 descend below its initial level?

I am not quite sure how to conclude solving this.

I have the tension of the string = (2*16*9.8)/(10) = 31.36

I have the acceleration = (+/-) (8-2)/(2+8)*g = 3/5*g = (+/-) 5.88

I just do not know how to go from here.

I know Vf^2=Vi^2+2*a*dy = Vf^2=2.8^2+2*(-5.88)*dy

dy is my answer, but then how do I get Vf^2 from what I know?

the velocity function is 5.88t-2.8

the position function is 2.94t^2-2.8t

K. I got it. VF^2=0

0-2.8^2=2*5.88*dy

-7.84=11.76dy

dy=-7.84/11.76

dy= -2/3