Wayne community college

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When a 0.101 kg mass is suspended at rest from a certain spring, the spring stretches 3.80 cm. Find the instantaneous acceleration of the mass when it is raised 5.90 cm, compressing the spring 2.10 cm.

  • Wayne community college -

    Yes

  • physics -

    w=kx
    .101g=k*3.80

    k=.101g/3.80

    force of compression=
    = (.101g/.0389)(.0590-.0380)
    figure that out, then
    using f=ma or a=f/m
    and we have to add g because the object is going down,

    a=g+(.101g/.0389)(.0590-.0380) /.101g)

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