Question

A 20-kg block is held at rest on the inclined slope by a peg. A 2-kg pendulum starts at rest in a horizontal position when it is released. The pendulum bob strikes the block at the bottom of its swing with a horizontal velocity of 15 m/s. The coefficient of restitution of the impact is e = 0.7, and the coefficient of kinetic friction between the block and the inclined surface is µk = 0.5. What distance does the block slide before stopping?

Answer 1

Complete Question

The diagram of this question is shown on the first uploaded image

Answer:

The distance the block slides before stopping is $d = 0.313 \ m$

Explanation:

The free body diagram for the diagram in the question is shown

From the diagram the angle is $\theta = 25 ^o$

         $sin \theta = \frac{h}{d}$

Where $h = h_b - h_a$

     So      $d sin \theta = h_b - h_a$

From the question we are told that

      The mass of the block is  $m = 20 \ kg$

       The mass of the pendulum is  $m_p = 2 \ kg$

       The velocity of the pendulum at the bottom of swing is $v_p = 15 m/s$

        The coefficient of restitution is  $e =0.7$

         The coefficient of kinetic friction is  $\mu _k = 0.5$

The velocity of the block after the impact is mathematically represented as

            $v_2 f = \frac{m_b - em_p}{m_b + m_p} * v_2 i + \frac{[1 + e] m_1}{m_1 + m_2 } v_p$

Where  $v_2 i$ is the velocity of the block  before collision which is  0

                  $= \frac{20 - (0.7 * 2)}{(2 + 20)} * 0 + \frac{(1 + 0.7) * 2 }{2 + 20} * 15$

Substituting value

                   $v_2 f = 2.310\ m/s$

According to conservation of energy principle

      The energy at point a  =  energy at point b

So    $PE_A + KE _A = PE_B + KE_B + E_F$

Where  

         $PE_A$ is the potential energy at A which is mathematically represented as

          $PE_A = m_b gh_a$ = 0 at the bottom

      $KE _A$ is the kinetic energy at A  which is mathematically represented as

               $K_A = \frac{1}{2} m_b * v_2f^2$                  

         $PE_B$ is the potential energy at B which is mathematically represented as  

            $PE_B = m_b gh$

From the diagram $h = h_b -h_a$

       $PE_B = m_b g(h_b - h_a)$

$KE _B$ is the kinetic energy at B  which is 0 (at the top )

Where is $E_F$ is the workdone against velocity  which from the diagram is

      $\mu_k m_b g cos 25 *d$

So

   $\frac{1}{2} m_b v_2 f^2 = m_b g h_b + \mu_k m_b g cos \25 * d$

Substituting values

   $\frac{1}{2} * 20 * 2.310^2 = 20 * 9.8 * d sin(25) + 0.5* 20 * 9.8 * cos 25 * d$    

So

       $d = 0.313 \ m$