A 1.50 kilogram, horizontal, uniform tray is attached to a vertical ideal spring

Question

A 1.50 kilogram, 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 that it can oscillate up and down. The tray is 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? (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 on the tray?

Answer is 0.094 meters for part (a)

Answer is .22 seconds for part (b)

Answer is 1.2 m/s for part (c)

Thanks!

Explanation / Answer

a )

x = m' g / k

= 9.8 X ( 1.5 + 0.275 ) / 185

x = 0.094 m

b )

T = 2 X 3.14 ( m ' / k )12

T = 0.6151 sec

9.4 = 15 cos ( 185 / ( 1.5 + 0.275 ) )1/2 t

t = 0.06721 sec

t' = T / 4 + t

= 0.6151 / 4 + 0.06721

t ' = 0.22 sec

c )

V = w ( A2 - x2 )1/2

V = ( 185 / 1.5 + 0.275 )1/2 ( 0.152 - 0.0942 )1/2

V =1.2 m/sec

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