Question:Block Ain Fig. P5.89 has mass , and block Bhas mass . The coefficient of kinetic friction between block B and the horizontal surface is . (a) What is the mass of block C if block Bis moving to the right and speeding up with an acceleration of ? (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 $m_A=4.00\text{kg}$.
• The mass of the block B is $m_B=12.00\text{kg}$.
• The kinetic friction coefficient between the horizontal surface and the block B is $\mu =0.25$.
• The acceleration of the block B is $a=2.00{\text{m/s}}^{\text{2}}$.

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:

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

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.

$\begin{array}{rcl}{T}_1& =& \left(4.00\text{kg}\right)\left(2.00{\text{m/s}}^{\text{2}}+9.8{\text{m/s}}^{\text{2}}\right)\\ & =& \left(4.00\text{kg}\right)\left(11.8{\text{m/s}}^{\text{2}}\right)\\ & =& 47.2\text{kg}·{\text{m/s}}^{\text{2}}\times \frac{1\text{N}}{1\text{kg}·{\text{m/s}}^{\text{2}}}\\ & =& 47.2\text{N}\end{array}$

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 N is applied along with the frictional force $f_k$.

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

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

$f_k=\mu m_{B}g$

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.

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

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

$$\begin{gathered} m_{C}g-T_2=m_{C}a \\ {m}_C\left(g-a\right)=T_2 \\ {m}_C=\frac{T_2}{g-a} \end{gathered}$$

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

Substitute the values in the above equation.

$\begin{array}{rcl}{m}_C& =& \frac{100.6\text{N}}{9.8{\text{m/s}}^{\text{2}}-2.0{\text{m/s}}^{\text{2}}}\\ & =& \frac{100.6\text{N}\times \frac{\text{1}\text{kg}·{\text{m/s}}^{\text{2}}}{\text{1}\text{N}}}{7.8{\text{m/s}}^{\text{2}}}\\ & =& \frac{100.6\text{kg}·{\text{m/s}}^{\text{2}}}{7.8{\text{m/s}}^{\text{2}}}\\ & =& 12.89\text{kg}\end{array}$

Thus, the mass of the block C is $12.89\text{kg}$.