Work done by spring with weight to the spring

[NoBodyKnows]

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?

Cheers!

*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

 

[PhanthomJay]

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.

 

[kerimek]

.*

Hello there,

Thank you for your question. You are correct in thinking that the weight of the spring needs to be considered when calculating the work done by the spring. The equation you have provided, Ws=-(.5kx^2)-m(spring)g, is the correct equation to use. The weight of the spring, m(spring)g, should be included in the calculation as it represents the force that the spring is exerting due to its own weight.

The reason for this is that when a spring is stretched or compressed, it is not just the external force that is doing work on it, but also the internal force of the spring itself. This internal force is equal to the weight of the spring, and it needs to be taken into account in the calculation of work done by the spring.

I hope this helps clarify your understanding. Keep up the good work!