# Boxes A and B have masses of 12.3 kg and 3.7 kg, respectively. The two boxes are released from...

## Question:

Boxes A and B have masses of 12.3 kg and 3.7 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.

## The energy lost in the friction:

The friction force acts in the opposite direction of the motion, therefore, friction dissipates the energy of the system or object while motion. It is directly proportional to the coefficient of friction of the surface.

Given data:

• Mass of the block A, {eq}\rm (m_{1}) = 12.3 \ kg {/eq}
• mass of the block B, {eq}\rm (m_{2}) = 3.7 \ kg {/eq}

Now considering the Initial energy of the system is zero.

Now, the final energy of the system would be

{eq}\rm E_{2} = m_{2}g(-d) + (f \times d) + \dfrac{1}{2}(m_{1} + m_{2})v^{2} \\ E_{2} = (-3.7 \times 9.8 \times 0.5) + (\mu \times m_{1}g \times d) + \dfrac{1}{2}(12.3 + 3.7) v^{2} \\ E_{2} = - 18.13 + 12.054 + 8v^{2} {/eq}

Now, applying the energy conservation

{eq}\begin{align} \rm E_{1} &= \rm E_{2} \\ 0 &=\rm - 18.13 + 12.054 + 8v^{2} \\ \rm 8v^2 &= 6.076 \\ \rm v &= \rm 0.87 \ m/s \\ \end{align} {/eq} 