Using Hooke's Law and Elastic Potential Energy equations

[mp9191]

I'm just having a bit of trouble understanding when to use F=kx and when to use W=.5kx^2 .
I understand that Hooke's law is for the spring force and elastic PE is obviously for work, but you could use either one of these equations to solve for, for example, k:

When solving for the spring constant k, you could use F=kx:
k = F/x

or, using elastic PE and the fact that W=F(cos0)x,
.5kx^2 = F(cos0)x
k= 2F/x

There is a flaw in my thinking somewhere but I'm just not seeing it. Could someone please help clarify this for me?

Thanks in advance!

 

[PhanthomJay]

I'm just having a bit of trouble understanding when to use F=kx and when to use W=.5kx^2 .
I understand that Hooke's law is for the spring force and elastic PE is obviously for work, but you could use either one of these equations to solve for, for example, k:

When solving for the spring constant k, you could use F=kx:
k = F/x

or, using elastic PE and the fact that W=F(cos0)x,
.5kx^2 = F(cos0)x
k= 2F/x

There is a flaw in my thinking somewhere but I'm just not seeing it. Could someone please help clarify this for me?

Thanks in advance!

That is a keen observation you have made, even though the 2nd result is incorrect. It shows nonetheless that you are putting forth a good effort to understand the principles. The reason why the 2nd result is wrong is because you assumed the spring force was constant. The equatioin W =Fxcos theta is valid for constant force. The spring force kx is not constant, it varies linearly as a function of x. So the work done is the average force of the spring times the distance. What is the average force of the spring as it moves from 0 to x? Or, if you like, you can use the calculus to solve for the work, W = _____?

 

[mp9191]

That is a keen observation you have made, even though the 2nd result is incorrect. It shows nonetheless that you are putting forth a good effort to understand the principles. The reason why the 2nd result is wrong is because you assumed the spring force was constant. The equatioin W =Fxcos theta is valid for constant force. The spring force kx is not constant, it varies linearly as a function of x. So the work done is the average force of the spring times the distance. What is the average force of the spring as it moves from 0 to x? Or, if you like, you can use the calculus to solve for the work, W = _____?

Ah, I see. I was overlooking the fact that W=Fxcos theta is only valid for constant forces. Your explanation was very clear. Thank you for the help!