Hookes Law and a uniform bar

[Aqua Marine]

Homework Statement

A uniform bar of an iron is supported by a long, uniform hooke's law spring. The spring is cut in half and two pieces are used to support the same bar. If the whole spring stretched by 4.0cm, by how much would each half strech?

Homework Equations

f=Kchange in X

The Attempt at a Solution

 

[Doc Al]

Hint: How does the spring constant of each half-spring compare to the original spring constant? (Does it get harder or easier to stretch?)

 

[Aqua Marine]

it become harder to stretch...

 

[Aqua Marine]

I tried solving it and dats what i got:

Fnet= ma
Fs + fg = 0
KX - mg = 0
k = (mg)/ X
(10kg x 9.8m/s) / 0.04
= 2450

first spring = (2450 x 2)
= 4 900
Fa = (k1 + k2)
X= (mg/ k1 + k2)
(10kg x 9.8) / ( 4 900 N/m+4 900N/m)
0.01m= X

 

[Aqua Marine]

please let me know if i did it the right way or not...Thank You so Much!

 

[Doc Al]

I tried solving it and dats what i got:

Fnet= ma
Fs + fg = 0
KX - mg = 0
k = (mg)/ X
(10kg x 9.8m/s) / 0.04
= 2450

first spring = (2450 x 2)
= 4 900
Fa = (k1 + k2)
X= (mg/ k1 + k2)
(10kg x 9.8) / ( 4 900 N/m+4 900N/m)
0.01m= X

Looks good to me! You assumed a mass of 10 Kg for the iron bar (which was not given); but since the answer does not depend on the mass, that's an OK strategy.

But realize you can also solve this algebraically without assuming values:

For the whole spring:
W = KX (where W is the weight of the bar and X = 4 cm)

So K = W/X

For the half-springs:
k = 2K = 2W/X

Since each spring supports half the weight:
W/2 = kx = (2W/X)x

So x = X/4 = 1 cm