Spring with weight question / Potential energy of the spring

[masterchiefo]

Homework Statement

The spring of a dynamo-meter grows 20cm when one suspends a 4N on weight. What is the potential energy of the spring when one suspends a weight of 12N?

Homework Equations

Ue = 1/2*k*r²
Ug= m*g*h

The Attempt at a Solution

so to find k I do this step.
1/2*k*0.2² = 4*0.2
k=40N/M

Now that I have k I can find the potential energy of the spring with 12N
1/2*40*r² =12*r
I find r to be 0.6
now I do 12*0.6 = 7.2J to fnd the potential energy of the spring.

The real answer is 3.6J but I can't figure out how to get this.

 

[PhanthomJay]

Looks like you tried to set the PE of the spring equal to the work done by the weight force, but you calculated the work wrong when doing it this way. Try Hookes law to calculate k.

 

[masterchiefo]

Looks like you tried to set the PE of the spring equal to the work done by the weight force, but you calculated the work wrong when doing it this way. Try Hookes law to calculate k.

I don't understand, I am already using Hookes law to calculate K.

How am I supposed to find K with only this formula W= 1/2*k*r²
I don't have K nor W and only r

 

[PhanthomJay]

I don't understand, I am already using Hookes law to calculate K.

How am I supposed to find K with only this formula W= 1/2*k*r²
I don't have K nor W and only r

Write down Hookes law. Solve for k knowing that the spring extends 20 cm in the equilibrium position when the weight is 4 N.

 

[masterchiefo]

Write down Hookes law. Solve for k knowing that the spring extends 20 cm in the equilibrium position when the weight is 4 N.

Thats Hookes law... W= 1/2*k*r2
4*0.2= 1/2*k*0.2²

 

[PhanthomJay]

Thats Hookes law... W= 1/2*k*r2
4*0.2= 1/2*k*0.2²

No that is not Hookes law. The work done by a spring is the negative of its change in PE which somewhat resembles your equation, but what you want is Hookes law to make life easier, please look it up.

 

[collinsmark]

As PhanthomJay said, W = (1/2)kr² is not Hooke's law. You don't want to use W = (1/2)kr² when calculating the spring constant. (Although you will use it later when calculating the spring's potential energy.)

If you are puzzled as to why this conservation of energy approach does not work for calculating the spring constant k, I might be of help. Using the equation
mgh = (1/2)kh²
Implies the following: Initially, the weight is put on the relaxed spring, then released such that weight falls on its own accord. Eventually, the weight's velocity will momentarily drop back to zero when it falls a distance h below its initial position. At this point in time the initial energy of mgh has been transferred to the potential energy of the spring, (1/2)kh².

But it won't stop there! The mass will then shoot back up to its initial position and oscillate back and forth like that until frictional forces slowly dampen the oscillation.

So that h that I describe above is not the h that you are looking for.

In this problem, instead of letting the mass fall on its own accord, the mass is gently lowered until it reaches a distance of 20 cm, at which point the system stays in a state of equilibrium (no oscillations). That's where you want to use Hooke's law to find the spring constant. Hooke's law describes force; not potential energy (at least not directly).