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.
[tex]m_{B}g - T = m_{B}a[/tex]
T = [tex]m_{B}g - m_{B}a[/tex] .......... (1)
Now, force of equation for box A is as follows.
[tex]T - f_{k, A} = m_{A}a[/tex]
[tex]T - \mu_{k}m_{A}g = ma[/tex]
T = [tex]\mu_{k}m_{A}g + m_{A}a[/tex] ........... (2)
We will equate both equation (1) and (2) as follows.
[tex]m_{B}g - m_{B}a[/tex] = [tex]\mu_{k}m_{A}g + m_{A}a[/tex]
[tex]m_{B}g - m_{B}a = \mu_{k}m_{A}g + m_{A}a[/tex]
[tex]m_{B}g - \mu_{k}m_{A}g = a(m_{A} + m_{B})[/tex]
a = [tex]\frac{4.8 kg - (0.20)(13.4 kg)}{13.4 kg + 4.8 kg} \times 9.8 m/s^{2}[/tex]
= 1.1415 [tex]m/s^{2}[/tex]
Now, we will calculate the velocity of the box after falling through a distance of 0.50 m as follows.
[tex]v_{B} = \sqrt{2aS}[/tex]
= [tex]\sqrt{(2 \times 1.1415 \times 0.50)}[/tex]
= 1.07 m/s
Thus, we can conclude that speed of the given boxes is 1.07 m/s.
