Question

What would be the speed of the boxes when box b has fallen a distance of 0.50 m? The coefficient of kinetic friction between box a and the surface it slides on is 0.20. Use conservation of energy.

Answer 1

The given question is incomplete. The complete question is as follows.

Boxes A and B in the figure have masses of 13.4 kg. and 4.8 kg. , respectively. The two boxes are released from rest. What would be the speed of the boxes when box b has fallen a distance of 0.50 m? The coefficient of kinetic friction between box a and the surface it slides on is 0.20. Use conservation of energy.

Explanation:

Let us assume that box B is falling down, so equation of force for box B is as follows.

         $m_{B}g - T = m_{B}a$

             T = $m_{B}g - m_{B}a$ .......... (1)

Now, force of equation for box A is as follows.

         $T - f_{k, A} = m_{A}a$

            $T - \mu_{k}m_{A}g = ma$

            T = $\mu_{k}m_{A}g + m_{A}a$ ........... (2)

We will equate both equation (1) and (2) as follows.

       $m_{B}g - m_{B}a$ = $\mu_{k}m_{A}g + m_{A}a$

   $m_{B}g - m_{B}a = \mu_{k}m_{A}g + m_{A}a$

    $m_{B}g - \mu_{k}m_{A}g = a(m_{A} + m_{B})$

      a = $\frac{4.8 kg - (0.20)(13.4 kg)}{13.4 kg + 4.8 kg} \times 9.8 m/s^{2}$

         = 1.1415 $m/s^{2}$

Now, we will calculate the velocity of the box after falling through a distance of 0.50 m as follows.

          $v_{B} = \sqrt{2aS}$

                    = $\sqrt{(2 \times 1.1415 \times 0.50)}$

                    = 1.07 m/s

Thus, we can conclude that speed of the given boxes is 1.07 m/s.