;;;;;Two springs are hanging from a ceiling in series. Spring 2 is closest to ce

Question

;;;;;Two springs are hanging from a ceiling in series. Spring 2 is closest to ceiling and has a spring constant of 12 N/m. It equilibrium position is at the 15cm mark. Spring 1 hangs from Spring 2 on the bottom and has spring constant of 8 N/m and it's equilibrium position is at the 25 cm marker. "An empty mass hanger hangs at the bottom of spring 1 A 100 g mass is placed on the mass hanger and the springs stretch to a new equilibrium position Determine ythe new equilibrium posistions for the two springs with the 100 gram mass on the hanger

Explanation / Answer

Here ,
when the 100 gm mass hanger is attached

mass , m = 0.100 Kg

for the spring 2 ,

change in length is x2

k * x2 = m * g

12 * x2 = 0.100 * 9.8

x2 = 0.0817 m = 8.17 cm

Now , for the spring 1 ,

change in length is x1

k * x1 = m * g

8 * x1 = 0.100 * 9.8

x1 = 0.1225 m

x1 = 12.25 cm

Now , for the spring 2 , equilibirum position of spring 2 = 15 + 8.17

equilibirum position of spring 2 = 23.17 cm

the equilibirum position of spring 2 is 23.17 cm

for spring 1 ,

equilibirum position of spring 1 = 25 + 8.17 + 12.25

equilibirum position of spring 1 = 35.42 cm

the equilibirum position of spring 1 is 35.42 cm

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