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Force on a spring Watch

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    If you pull a spring with a mass attached downwards and release it, why is mass x acceleration only equal to the tension and not the tension subtract the weight?

    I'm a little confused

    Thanks!
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    (Original post by PhyM23)
    If you pull a spring with a mass attached downwards and release it, why is mass x acceleration only equal to the tension and not the tension subtract the weight?

    I'm a little confused

    Thanks!
    Yeah, it should be T-mg=ma. Perhaps the mass spring system is assumed to be in the absence of gravitational fields.
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    (Original post by Absent Agent)
    Yeah, it should be T-mg=ma. Perhaps the mass spring system is assumed to be in the absence of gravitational fields.
    Thanks for the reply!

    In the AQA A2 Physics textbook, it says the acceleration x mass = restoring force which is equal to kx. Isn't kx just the tension in the spring. If so, why is it kx = ma and not kx-mg=ma?
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    (Original post by PhyM23)
    Thanks for the reply!

    In the AQA A2 Physics textbook, it says the acceleration x mass = restoring force which is equal to kx. Isn't kx just the tension in the spring. If so, why is it kx = ma and not kx-mg=ma?
    I think you need to be very clear whether you're measuring x from the unstretched length of the spring or measuring x from the equilibrium point of the spring supporting the weight of the mass.

    probably you want to be thinking about x measured from the equilibrium position, is there a diagram or something?
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    (Original post by PhyM23)
    Thanks for the reply!

    In the AQA A2 Physics textbook, it says the acceleration x mass = restoring force which is equal to kx. Isn't kx just the tension in the spring. If so, why is it kx = ma and not kx-mg=ma?
    I don't think that's correct. Only the resultant force on the mass causes it to accelerate so it cannot be equal to the tension of the string. However, it is the case that the tension in the string is proportional to its extension, that is; T=kx
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    Thank you both for replying, I've managed to understand it now
 
 
 
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