Question

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

Answer 1

Complete Question

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

Answer:

The velocity is  $=4.51m/s$  

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

                      $U = \frac{1}{2} kx^2$

Where x  is the length the spring is displaced

k is the force constant of the string

         $U = \frac{1}{2} * 627 * (0.6)^2$

          $= 112.86 J$

   Now referring to the formula above

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

                $\frac{1}{2} mv^2 = 112.86 - \mu_kmgx$

                $v^2 = \frac{2(112.86 -\mu_kmgx)}{m}$

                $v = \sqrt{\frac{2(112,86- \mu_kmgx)}{m}}$

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

  Therefore   $v = \sqrt{\frac{2(112.86- (0.4 *9*9.8*0.6))}{9} }$

                        $=4.51m/s$