Work done by spring with weight to the spring

AI Thread Summary
The work done by a spring is typically calculated using the formula Ws=-0.5kx^2. When considering a spring with mass, the weight of the spring itself can influence the total work done, particularly in vertical configurations. The equation may need to incorporate the spring's weight, represented as Ws=-(0.5kx^2)-m(spring)g, to account for gravitational effects. However, in horizontal setups, the mass of the spring does not significantly impact the work done. Overall, the mass of the spring is relevant in vertical applications but negligible in horizontal ones.
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Hey, I just have a quick question. I understand that the work done by a spring is considered as the following; Ws=-.5kx^2. What if the spring has weight to it? Would the work done by the spring be Ws=-(.5kx^2)-m(spring)g? Does the weight get added or is it not even considered in that equation?

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*i'm saying the weight is the spring, I understand where the weight the spring is holding is associated but what about the mass of the spring if there is one
 
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In the absence of more data, about all you can say about the work done by a spring that is not massless is, since both spring and gravity forces are conservative,
W_s + W_g = - (\Delta U_s + \Delta U_g). If the spring was horizontal, its mass wouldn't matter, but the case would be different if the spring was vertically mounted.
 
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