Question

Three identical 6.67-kg masses hang at rest from three identical springs, numbered as in the figure. Each spring has an unstretched length of 0.23 m and a spring constant of 8.13 kN/m. How long, L , is spring 2?
[Hint: Draw some FBDs and apply Newton's second law. Remember to account for both the spring's stretch and its unstretched length.]

Answer 1

The length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.

Extension of the spring 2

The extension of the spring 2 is determined from the net force acting on spring 2.

F(net) = Kx

Net force on spring 2 = sum of forces acting downwards - sum of forces acting upwards.

Net force on spring 2 = (force on 2 + force on 3) - (force on 1)

F(net) = (m₂ + m₃)g - (m₁)g

F(net) = (6.67 + 6.67)9.8  -  (6.67)9.8

F(net) = 130.732 - 65.366

F(net) = 65.366 N

x = F/k

x = (65.366)/(8130)

x = 0.00804 m

The length of spring 2 = x + 0.23 m

                                     = 0.00804 m + 0.23 m = 0.238 m

Thus, the length of the spring 2 due to arrangement of the equal masses is determined as 0.238 m.

Learn more about extension of elastic material here: brainly.com/question/26838571