- #1

Elias Waranoi

- 45

- 2

## Homework Statement

A 1.50-kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 185 N/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 to point A, which is 15.0 cm below the equilibrium point, and released from rest. (a) How high above point A will the tray be when the metal ball leaves the tray?

m

_{t}=1.5 kg

m

_{b}= 0.275 kg

k = 185 N/m

A = -0.15 m

## Homework Equations

F = -kx

F = ma

## The Attempt at a Solution

My attempt:

The maximum downward acceleration of the ball is -g = -9.8 so the ball should leave the tray when the acceleration of the tray is also -g.

F = -kx = m

_{t}a = -m

_{t}g

x = m

_{t}g/k = 0.0795 m

Δx = x - A = 0.229 m

This is wrong and I would have gotten the right answer if I used m

_{t}+ m

_{b}instead of just m

_{t}in the above equations but I can't understand why. When the ball leaves the tray the weight of the ball on the tray should be zero (because they're not in contact anymore) combined with the acceleration of the tray being -g should give the equation x = m

_{t}g/k. But apparently this logic is flawed since the right answer is given by x = (m

_{t}+ m

_{b})g/k

Please tell me what's wrong with my logic.