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

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To determine how high above point A the tray will be when the metal ball leaves, the normal force on the ball must be zero. The problem involves a 1.5 kg tray attached to a spring with a force constant of 185 N/m and a 275 g ball resting in it. The tray is initially pushed down 15 cm and released, leading to oscillation. The equations of motion involve the gravitational force on the ball and the spring force acting on the tray. Understanding the forces and the point at which the normal force becomes zero is crucial for solving the problem accurately.
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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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