A 1.70 kg , horizontal, uniform tray is attached to a vertical ideal spring of f

Question

A 1.70 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/mand a 280 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 15.3 cm below its equilibrium point (call this point A) and released from rest.

A.) How high above point A will the tray be when the metal ball leaves the tray? (Hint: This does not occur when the ball and tray reach their maximum speeds.)

B.) How much time elapses between releasing the system at point A and the ball leaving the tray?

C.) How fast is the ball moving just as it leaves the tray?

Explanation / Answer

given data,

m = 1.7 + 0.28

= 1.98 kg

k = 200 N/m

Amplitude, A = 15.3 cm = 0.153 m

A) It happens at h = 2*A

= 2*15.3

= 30.6 cm or 0.306 m

B) Time period of motion, T = sqrt(k/m)/(2*pi)

= sqrt(200/1.98)/(2*pi)

= 1.6 s

time taken form lower point to higher point, t = T/2

= 1.6/2

= 0.8 s

c) Angular frequency, w = sqrt(k/m)

= sqrt(200/1.98)

= 10.05 rad/s

Vmax = A*w

= 0.153*10.05

= 1.54 m/s

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