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