In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring constant for the left-hand spring is 2.7 N/cm and the spring constant for the righ-hand spring is 4.3 N/cm. How far down will the 37 N weight stretch the springs?

Question

Physics

Gauge Cox

5 May, 02:30

In the parallel spring system, the springs are positioned so that the 37 N weight stretches each spring equally. The spring constant for the left-hand spring is 2.7 N/cm and the spring constant for the righ-hand spring is 4.3 N/cm. How far down will the 37 N weight stretch the springs?

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Answers (1)

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Leyla · Answer 1

x = 5.29 m

Explanation:

given,

weight of stretch = 37 N

left-hand spring constant (k₁) = 2.7 N/cm

right hand spring constant (k₂) = 4.3 N / cm

spring are connected in parallel

F = F₁ + F₂

F = k₁x + k₂x

F = (k₁ + k₂) x

37 = (4.3 + 2.7) x

7 x = 37

x = 5.29 m