Three identical 3.40kg masses are hung by three identical springs, as shown in t

Question

Three identical 3.40kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 5.10kN/m and was 12.0cm long before any masses were attached to it. How long is each spring when hanging as shown?
(Hint: First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)(Figure 1)

Part A

bottom spring:

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Part B

middle spring:

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Part C

top spring:

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Three identical 3.40kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 5.10kN/m and was 12.0cm long before any masses were attached to it. How long is each spring when hanging as shown?
(Hint: First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)(Figure 1)

Part A

bottom spring:

l1 =   m  

SubmitMy AnswersGive Up

Part B

middle spring:

l2 =   m  

SubmitMy AnswersGive Up

Part C

top spring:

l3 =   m  

SubmitMy AnswersGive Up

Three identical 3.40kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 5.10kN/m and was 12.0cm long before any masses were attached to it. How long is each spring when hanging as shown? (Hint: First isolate only the bottom mass. Then treat the bottom two masses as a system. Finally, treat all three masses as a system.)(Figure 1) Part A bottom spring: l1 = m SubmitMy AnswersGive Up Part B middle spring: l2 = m SubmitMy AnswersGive Up Part C top spring: l3 = m SubmitMy AnswersGive Up

Explanation / Answer

start from the lowermost block

mg = kx1

x1 = 3.4*9.8/5100 = 0.006533 m        (elongation in lower spring)

now the middle block

mg + kx1 = kx2

x2 = 66.64/5100 = 0.0131 m     (elongation in middle spring)

now the upper block

mg + kx2 = kx3

x3 = 99.96/5100 = 0.0196 m

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