How can i solved this problem Three identical 4.20 k g masses are hung by three

Question

How can i solved this problem

Three identical   4.20 kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 6.30

kN/m and was 19.0cm 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.

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 Provide FeedbackContinue Figure 1 of 1 Three identical 4.20 kg masses are hung by three identical springs, as shown in the figure. Each spring has a force constant of 6.30 kN/m and was 19.0cm 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.

Explanation / Answer

weight of the mass = 4.20 kg*9.8 m/s^2 = 41.16

The spring will have to exert a force equal and opposite the weight of the mass. To do so it will be stretched according to

F = k*d
where F = 41.16
k = 6.30*10^3 N/m

41.16 = 6.30*10^3 * d
d = 6.53*10^-3 = 0.00653 m = 0.653 cm

Final elongation = 19+0.653 =19.653 cm

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