Three identical 5.20 kg masses are hung by three identical springs, as shown in

Question

Three identical 5.20 kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 8.30 kN/m and was 13.0 cm 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:

6.26

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Part B

middle spring:

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Part C

top spring:

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Figure 1 of 1

Three identical 5.20 kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 8.30 kN/m and was 13.0 cm 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 =

6.26

  m  

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Incorrect; Try Again; 4 attempts remaining

Part B

middle spring:

l2 =   m  

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Part C

top spring:

l3 =   m  

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Figure 1 of 1

Explanation / Answer

m = 5.2 kg

k = 8.3 kN/m

Bottom spring

mass supported = 5.2 kg

Force = mg = 5.2*9.81 = 51.012 N

Extension = F/k = 51.012/8300 = 0.006146 m

So length of bottom spring = 0.13 + 0.006146 = 0.136146 m

Middle spring

Mass supported = 5.2+5.2 = 10.4 kg

Force = 10.4*9.81 = 102.024 N

Extension = F/k = 102.024/8300 = 0.01229 m

So length of middle spring = 0.13+0.01229 = 0.14229 m

Top spring

mass supported = 5.2+5.2+5.2 = 15.6 Kg

Force = 15.6*9.81 = 153.036 N

Extension = F/k = 153.036/8300 = 0.018438 m

Length of top spring = 0.13+0.018438 = 0.148438 m

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