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anastassius [24]
3 years ago
5

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.

Physics
1 answer:
bija089 [108]3 years ago
5 0

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.

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