A 1.30 kg , horizontal, uniform tray is attached to a vertical ideal spring of f
ID: 1472584 • Letter: A
Question
A 1.30 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/mand a 275 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 13.3 cm below its equilibrium point (call this point A) and released from rest. A)How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.) B)How much time elapses between releasing the system at point A and the ball leaving the tray? C)How fast is the ball moving just as it leaves the tray? Please I need the answers .
Explanation / Answer
A)
at a height of y = 2A = 2*13.3 = 26.6 cm
B)
time period = T = 2*pi*sqrt(m/K)
T = 2*pi*sqrt((1.3+0.275))/200) = 0.56 s
time = half of the time period = 0.28 s
C)
change in potential energy of spring = 2*0.5*k*x^2
increase gravitational potential energy of ball = 0.275*9.8*2*0.133
final KE = 0.5*m*v^2 = 2*0.5*k*x^2 - 0.275*9.8*2*0.133
0.5*0.275*v^2 = (2*).5*200*0.133^2) - (0.275*9.82*0.133)
v = 4.81 m/s
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