Question

A 1.70 kg, horizontal, uniform tray is attached to a vertical ideal spring of force constant 195 N/m and a 285 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 15.6 cm below its equilibrium point (call this point A) and released from rest.

How much time elapses between releasing the system at point AAA and the ball leaving the tray? Express your answer with the appropriate units.

t =___

Answer 1

The time elapsed between releasing the system at point A and the ball leaving the tray is 0.414 seconds.

Step-by-step explanation:

To find the time it takes for the ball to leave the tray, we need to first determine the time period of the oscillations of the tray-spring system.

We can use the equation for the time period of a mass-spring system: T = 2π√(m/k), where T is the time period, m is the mass of the ball, and k is the force constant of the spring. Plugging in the values, we have T = 2π√(0.285 kg / 195 N/m). Solving this equation gives us T = 0.827 s.

Since the ball is released at point A, we need to find the time it takes for the tray to reach the equilibrium point before the ball leaves the tray.

This is half of the time period, so t = T/2 = 0.827 s / 2 = 0.414 s.

Therefore, the time elapsed between releasing the system at point A and the ball leaving the tray is 0.414 seconds.