The 9 kg block is then released and accelerates to the right, toward the 5 kg block. The surface is rough and the coefficient of friction between each block and the surface is 0.4 . The two blocks collide, stick together, and move to the right. Remember that the spring is not attached to the 9 kg block. Find the speed of the 9 kg block just before it collides with the 5 kg block. Answer in units of m/s

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Complete Question

The complete question is shown on the first and second uploaded image

Answer:

The velocity is  [tex]=4.51m/s[/tex]  

Explanation:

The kinetic energy of the 9 kg can be determined by these expression

        Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

  Now to obtain the initial energy stored

               Let U denote the initial energy stored and

                      [tex]U = \frac{1}{2} kx^2[/tex]

Where x  is the length the spring is displaced

k is the force constant of the string

         [tex]U = \frac{1}{2} * 627 * (0.6)^2[/tex]

          [tex]= 112.86 J[/tex]

   Now referring to the formula above

i.e          Kinetic energy of 9 kg  block = initial energy stored - energy lost as a result of friction

                [tex]\frac{1}{2} mv^2 = 112.86 - \mu_kmgx[/tex]

                [tex]v^2 = \frac{2(112.86 -\mu_kmgx)}{m}[/tex]

                [tex]v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}[/tex]

and we are told that coefficient of friction  = 0.4 and the mass is 9 kg ,the acceleration due to gravity [tex]= 9.8m/s^2[/tex]  this displacement length of spring = 0.6

  Therefore   [tex]v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }[/tex]

                        [tex]=4.51m/s[/tex]      

           

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Ver imagen okpalawalter8