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g A cylinder of mass m is free to slide in a vertical tube. The kinetic friction force between the cylinder and the walls of the tube has magnitude f. You attach the upper end of a lightweight vertical spring of force constant k to the cap at the top of the tube, and attach the lower end of the spring to the top of the cylinder. Initially the cylinder is at rest and the spring is relaxed. You then release the cylinder. What vertical distance will the cylinder descend before it comes momentarily to rest

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okpalawalter8

Answer:

The vertical distance is  [tex]d = \frac{2}{k} *[mg + f][/tex]

Explanation:

From the question we are told that

   The mass of the cylinder is  m

    The kinetic frictional force is  f

Generally from the work energy theorem

    [tex]E  =  P +  W_f[/tex]

Here E the the energy of the spring which is increasing and this is mathematically represented as

       [tex]E =  \frac{1}{2} * k  *  d^2[/tex]

Here k is the spring constant

        P is the potential energy of the cylinder which is mathematically represented as

     [tex]P  = mgd[/tex]

And

     [tex]W_f[/tex]  is the workdone by friction which is mathematically represented as

      [tex]W_f  =  f *  d[/tex]

So

    [tex] \frac{1}{2} * k  *  d^2 =  mgd +  f *  d [/tex]

=>    [tex] \frac{1}{2} * k  *  d^2 =  d[mg +  f    ][/tex]

=>  [tex] \frac{1}{2} * k  *  d =  [mg +  f    ][/tex]

=> [tex]d = \frac{2}{k} *[mg + f][/tex]