Usern4me said:
Homework Statement
A 1.60 kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/m and a .3 kg 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 .149 meters below its equilibrium point to a point A and released from rest. At approximately .242 meters above point A the ball will lose contact with the spring, find the time this happens and the velocity of the ball at that time.
Homework Equations
x=Acos(w*t)
Fx=-200*x-9.8
The Attempt at a Solution
There was an earlier part of the question where you were supposed to find the height at which the ball leaves the tray and I thought that I could derive the time from that equation but I wasn't able to.
As for the velocity I tried taking the second integral of the acceleration, but that also led me to the wrong answer.
OK to give you a guide as to what is going on.
Suppose we just had the an empty tray. Suppose also that the weight of the tray would compress the spring say 5cm.
The most common application we come across is to press that tray down a further, perhaps, 3 cm and release it.
When released, the tray will then oscillate with SHM with an amplitude 3 cm.
Suppose instead, that tray was pressed down not just an extra 3 cm, but an extra 10 cm.
The oscillation would now be very violent, and if we had failed to attach the tray to the spring, it would fly off into the air.
At its highest point, the [now stretched] spring would be pulling the tray down. The combined effect of the weight of the tray, plus the pull from the spring may supply a force of sufficient size that the tray is accelerating down at a rate higher than g [9.8].
If that were the case, and the tray had an unattached ball on it, the ball would leave the tray, since it would be accelerating down at only 9.8, like any object in free fall.
In this problem you have to find where the tray is when that condition is reached, and how fast the tray is traveling at the time.
