A 1.50 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 180 N/m and a 260 g metal ball is in the tray. The spring is below the tray, so it can oscillate up-and-down. The tray is then pushed down 15.2 cm below its equilibrium point (call this point A) and released from rest.
How fast is the ball moving
just as it leaves the tray? (Answer is NOT 1.67 m/s or 1.98
m/s)
Initial Energy stored in spring = 0.5*180*0.152 = 13.68 J
at h = 24.8 cm
PE gained = (1.5+0.26)*9.8*0.248 = 4.277 J
KE gained = 0.5(1.5+0.26)v^2 = 0.88v^2
By energy balance
0.88v^2 + 4.277 + 0.5*180*(0.248 - .152)^2 = 13.68
v = 3.121 m/s
A 1.50 kg , horizontal, uniform tray is attached to a vertical ideal spring of force...
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