A 1.80 kg , horizontal, uniform tray is attached to a vertical ideal spring of f

Question

A 1.80 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/m and a 255 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 15.7 cm below its equilibrium point (call this point A) and released from rest.

Part 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.)

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Part B

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

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Part C

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

A 1.80 kg , horizontal, uniform tray is attached to a vertical ideal spring of force constant 200 N/m and a 255 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 15.7 cm below its equilibrium point (call this point A) and released from rest.

Part 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.)

h =   cm  

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Part B

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

t =   s  

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Part C

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

v =   m/s  

Explanation / Answer

mg = kx

x = mg/k

m = 1.8 + 0.255 kg = 2.055 kg

k = 200 N/m

x = 0.1 m = 10 cm

this point is 10 cm above the equilibrium point

so 10 + 15.7 = 25.7 cm above the point A

part b )

w = sqrt(k/m) = 9.865 rad/s

this point is -10 cm above the equilibrium point

x = Acoswt

wt = cos^-1(x/A)

A = 15.7 cm

wt = 2.267 rad

t = 2.267/w = 0.2298 s = 0.23 s

part c )

kA^2/2 = kx^2/2 + mv^2/2

v = sqrt(k/m(A^2-x^2))

v = 1.188 m/s = 1.2 m/s

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