Q 89P - University Physics with Modern Physics

Question

Question: Block A in Fig. P5.89 has mass , and block B has mass . The coefficient of kinetic friction between block B and the horizontal surface is . (a) What is the mass of block C if block B is 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?

Step-by-Step Solution

Answer

(a) The mass of the block C is 12.89kg.

(b) The tensions in each cord are 100.6N and 47.2N respectively.

Step 1: Identification of the given data

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 $a=2.00 {\text{m/s}}^{\text{2}}$.

Step 2: Significance of the tension

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.

Step 3: (b) Determination of the tension in the cords

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.

Step 4: (a) Determination of the mass of the block C

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 $12.89 \text{kg}$.