Solving Spring Tray Problem: Height Above Point A When Ball Leaves Tray

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SUMMARY

The discussion focuses on determining the height above point A that a tray will rise when a metal ball leaves it. The tray, weighing 1.5 kg, is attached to a spring with a force constant of 185 N/m and is initially displaced 15 cm below its equilibrium position. The key equations involved are the force equation f = -kx and the energy equation E = 0.5 mv^2 + 0.5 k A^2. The critical point for the ball leaving the tray occurs when the normal force on the ball becomes zero, indicating that the gravitational force equals the spring force.

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A 1.5 kg horizontaal uniform tray is attached to a vertical ideal spring of force constant 185N/m and 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 15 cm below its equilibrium point called point A and released from rest.

How high above point A will the tray be when the metal ball leaves the tray.

Equations

f=-kx
E=.5 mv^2+.5 k A^2

Im thinking that this will occur when the normal force on the ball is zero.


so i set up the mg=-kx and solved for x but that didnt lead me too the right answer.

Any advice would be greatly appreciated.
 
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Punkyc7 said:
Im thinking that this will occur when the normal force on the ball is zero.
And when is the normal force zero on the ball? What forces act on the ball? What is its acceleration till it moves together with the tray?

ehild
 

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