Block Ain Fig. P5.89 has mass 4.00 kg, and block Bhas mass 12.00 kg. The coefficient of kinetic friction between block B and the horizontal surface is 0.25. (a) What is the mass of block C if block Bis moving to the right and speeding up with an acceleration of $\mathbf{2}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$? (b) What is the tension in each cord when block B has this acceleration?

The given data can be listed below as:

• The mass of the block A is $m_a=4.00kg$.
• The mass of the block B is $m_b=12.00kg$.
• The kinetic friction coefficient between the horizontal surface and the block B is $\mu =0.25$.
• The acceleration of the block B is $\mathrm{a}=2.00\mathrm{m}/{\mathrm{s}}^2$.

The tension is described as the force that is mainly transmitted in an axial direction with the help of a string. Moreover, tension is also the pair of action and reaction forces that acts on an object.

The free body diagram of the system has been drawn below:

Here, in the diagram, it has been identified that the tensions$T_1$ and$T_2$ are mainly acting on the system.

From the above diagram, the equation of the tension between the block A and block B is expressed as:

$\begin{array}{l}{T}_1-m_{A}g=m_{A}a\\ T_1=m_{A}g+m_{A}a\end{array}$

Here, $T_1$is tension between the block A and block B, $m_A$is the mass of the block A, a is the acceleration of the block B and g is the acceleration due to gravity.

Substitute the values in the above equation.

$$\begin{gathered} T_1=\left(4.00\mathrm{kg}\right)\left(2.00\mathrm{m}/{\mathrm{s}}^2+9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =\left(4.00\mathrm{kg}\right)\left(11.8\mathrm{m}/{\mathrm{s}}^2\right) \\ =47.2\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2} \\ =47.2\mathrm{N} \end{gathered}$$

The free body diagram of the block B has been drawn below:

Here, in the diagram, it has been identified that apart from the tensions, a normal force is applied along with the frictional force $f_k$.

The equation of the tension between the block B and C is expressed as:

$$\begin{gathered} T_2-f_k-T_1=m_{B}a \\ \begin{array}{cc}{T}_2=m_{B}a+f_k+T_1& \end{array} \end{gathered}$$ …(i)

Here, $T_2$is the tension between the block B and C, $m_B$is the mass of the block B and $f_k$is the frictional force.

The equation of the frictional force is expressed as:

$f=\mu m_{B}g$

Substitute$\mu m_{B}g$ for$f_k$ in the above equation.

$T_2=m_{B}a+\mu m_{B}g+T_1$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{T}}_2=(12.00\mathrm{kg})\left(2.00\mathrm{m}/{\mathrm{s}}^2\right)+\left(0.25\right)\left(12.00\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)+47.2\mathrm{N} \\ =(24.00\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2)+\left(29.4\mathrm{kg}·\mathrm{m}/{\mathrm{s}}^2\right)+47.2\mathrm{N} \\ =\left(53.4\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2\times \frac{1\mathrm{N}}{1\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2}\right)+47.2\mathrm{N} \\ =100.6\mathrm{N} \end{gathered}$$

Thus, the tensions in each cord are 100.6 N and 47.2 N respectively.

The free body diagram of the block C has been drawn below:

In the above diagram, both the tension$T_2$ and the weight of the block that is the product of the mass of the block and acceleration due to gravity$m_{c}g$ is acting.

The equation of the mass of the block C is expressed as:

$\begin{array}{l}{m}_{c}g-T_2=m_{c}a\\ m_c\left(g-a\right)=T_2\\ m_c=\frac{T_2}{g-a}\end{array}$

Here,$m_c$ is the mass of the block C.

Substitute the values in the above equation.

$$\begin{gathered} m_c=\frac{100.6\mathrm{N}}{9.8\mathrm{m}/{\mathrm{s}}^2-2.0\mathrm{m}/{\mathrm{s}}^2} \\ =\frac{100.6\mathrm{N}\times \frac{1\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2}{1\mathrm{N}}}{7.8\mathrm{m}/{\mathrm{s}}^2} \\ =\frac{100.6\mathrm{kg}\cdot \mathrm{m}/{\mathrm{s}}^2}{7.8\mathrm{m}/{\mathrm{s}}^2} \\ =12.89kg \end{gathered}$$

Thus, the mass of the block C is 12.89 kg.