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A 0.45kg mass is suspended on a spring that stretches 4.0cm . Part A - what is t

ID: 1292574 • Letter: A

Question

A 0.45kg mass is suspended on a spring that stretches 4.0cm .

Part A - what is the spring constant?
Part B - What added mass would stretch the spring an additional 1.0cm ?
Part C - What is the change in potential energy when the mass is added? A 0.45kg mass is suspended on a spring that stretches 4.0cm .

Part A - what is the spring constant?
Part B - What added mass would stretch the spring an additional 1.0cm ?
Part C - What is the change in potential energy when the mass is added?
Part A - what is the spring constant?
Part B - What added mass would stretch the spring an additional 1.0cm ?
Part C - What is the change in potential energy when the mass is added?

Explanation / Answer

A)

F = k x

m g = k x

k = mg/ x

k = ( 0.45 * 9.8) / ( 0.04) = 110.25 N /m

B) For extra 1 cm stretching

x = 0.05 m

m * 9.8 = 110.25 * 0.05

m = 0.5625 kg

Extra mass = 0.5625 - 0.45 = 0.1125 kg = 112.5 grams

C) Change in Potential energy = 1/2 * k ( x1^2 - x2^2 )

= 1/2 * 110.25 * ( 0.05^2 - 0.04^2)

= 0.0496 J