Work done to stretch a spring vs Work done by a spring?

[lu6cifer]

Why is it that the equation for work done to stretch a spring is
W = 1/2k(x₂²-x₁²)

while the work done by a spring on an object is
W = 1/2k(x₁²-x₂²) ?

That is, how come it's x₂²-x₁² for the first case, and x₁²-x₂² for the second?

 

[Vanadium 50]

In one case it's the work done on the spring and in the other its work done by the spring.

 

[lu6cifer]

No, I get the semantics part, that in one case it's work done on a spring and in another it's work that's being done by the spring, but what's the physics explanation for why the x's are positioned where they are?

 

[radou]

The works done are equal, but opposite.

 

[lu6cifer]

I realize that they're opposites, meaning one is a positive value and the other is negative, but how is that determined? That is, how would you derive that based on the laws of physics?

 

[radou]

I realize that they're opposites, meaning one is a positive value and the other is negative, but how is that determined? That is, how would you derive that based on the laws of physics?

If a spring is stretched by a force, there are internal forces acting on the deformations of the spring, in a manner resisting the external applied force which stretches the spring, and their work is opposite of the work done by the external force. If the spring stretches, intuitively you could say that the particles in the spring tend to move apart, so between them there act forces which "try to move them back together". Fundamentally, it's the principle of action and reaction.

 

[russ_watters]

I realize that they're opposites, meaning one is a positive value and the other is negative, but how is that determined? That is, how would you derive that based on the laws of physics?

Displacement is calculated by convention to be positive.

Draw a set of axes. If you start at the origin and move to the right 5 units, your displacement is (x2-x1)=5-0=5. If you start at the origin and move to the left, your displacement is (x1-x2)=0-(-5)=5.

 

[sophiecentaur]

I think that the confusion arises because x1 and x2 haven't actually been defined here. One x is the shorter length and the other is the longer length. It would be better to state that you put the 'start' value and 'finish' value in the same place in the formula every time. This will produce a sign for the value of work which tells you whether it's put in or got out.
The two quoted versions of the formulae are 'verbal' ways of indicating the sign of the answer.
I hope that hasn't just added confusion!

 

[sophiecentaur]

The 'sign' of the work refers to whether you put energy in or get it out.

 

[Doc Al]

Realize that you are replying to a thread almost 9 months old.

 

[sophiecentaur]

Durr!
Will try harder next time.

 

[woodie37]

When an object does work on another object, it applies a force for a distance (W=Fd). Initially the work done or energy stored on a spring is (1/2)k(x1)^2 and the final is (1/2)k(x2)^2, which results in W=(1/2)k((x2)^2-(x1)^2). Now, when the spring releases, the initial position is x2 while the final position is x1, thus the order of the 2 terms in brackets become reversed. It can also be thought of as negative work done on the object. (-(a-b)=b-a) where a is x1 and b is x2