Question

State Hooke's law. A spring will stretch 12 cm when a 250-g mass is hung on it. If two springs are hung side by side and both support the 250-g mass, how much will each spring stretch? Explain.

Answer 1

When two springs are side by side like you describe , each spring will stretch the same amount, x, as given by the equation:

|Fs| = k1*x+k2*x, or
|Fs| = (k1+k2)*x

The mass is motionless, so |Fs| = mg

mg = (k1+k2)*x

x = mg/(k1+k2)

The length of stretch depends on the spring constant of each spring.

Answer 2

How do I find out the spring constant?

Answer 3

Since "A spring will stretch 12 cm when a 250-g mass is hung on it,"

Fs = kx --> mg = kx --> k = mg/x --> k=(.250 kg)(9.8 N/kg )/(12cm) = .204 N/cm

I guess the question wants to assume the two springs are equal, so

x = mg/(2k) --> x (two springs) = .250kg*9.8N/kg/[2*.204 N/cm] = 6 cm

The elongation length halves.

Answer 4

I honestly am not comprehending what I'm seeing. Not at all.

Answer 5

Force exerted by the spring = Fs
k = spring constant
x = length of stretch

Fs=kx

rearrange to put the focus on length of stretch:

x = Fs/k

In our case,
Fs = weight of the "mass", and that stays the same.

So the question is basically asking -- if Fs is constant, how does doubling k, the spring constant, affect the length of stretch?

if k doubles and Fs remains the same...

x = Fs/(2k)

Comparing old x and new x:

xbefore:xafter
Fs/k:Fs/(2k)
1:1/2

Therefore the same weight stretches the spring half as much when the weight is suspended on two springs side by side.

Answer 6

I'm giving up... /: thank you though

Answer 7

Simple 2 springs at 12 cm is 2x12 = 24 cm