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masha68 [24]
3 years ago
5

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
Physics
1 answer:
sdas [7]3 years ago
5 0

Answer:

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

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

    E  =  P +  W_f

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

       E =  \frac{1}{2} * k  *  d^2

Here k is the spring constant

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

     P  = mgd

And

     W_f  is the workdone by friction which is mathematically represented as

      W_f  =  f *  d

So

    \frac{1}{2} * k  *  d^2 =  mgd +  f *  d

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

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

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

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