Question

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?

Answer 1

x = 5.29 m

Step-by-step 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