Chapter 4, Problem

(a)

To determine

The net force acting on the entire two-block system.

Answer

The net force acting on the entire two-block system is $16\text{N}$.

Explanation of Solution

Given info: The horizontal force acting on $4\text{kg}$ mass along the right direction is $30\text{N}$, that along the left direction is $8\text{N}$ and horizontal force acting on $2\text{kg}$ mass along left direction is $6\text{N}$.

Write the expression for the net horizontal force.

$F_{\text{net}}=F_{\text{right}}-F_{\text{left}}$

Here,

$F_{\text{net }}$ is the net force acting on the system

$F_{\text{right}}$ is the horizontal force acting in the right direction

$F_{\text{left}}$ is the horizontal force acting in the left direction

Total force on the left direction is the sum of $6\text{N}$ and $8\text{N}$.

That is,

$\begin{array}{c}{F}_{\text{left}}=6\text{N+8}\text{N}\\ \text{=14}\text{N}\end{array}$

Substitute $\text{14}\text{N}$ for $F_{\text{left}}$ and $30\text{N}$ for $F_{\text{right}}$ in the above equation to get $F_{\text{net}}$.

$\begin{array}{c}{F}_{\text{net}}=30\text{N}-\text{14}\text{N}\\ =16\text{N}\end{array}$

Conclusion:

Thus, the net force acting on the entire two-block system is $16\text{N}$.

(b)

To determine

The acceleration of the system.

Answer

The acceleration of the system is $2.67\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The masses of the blocks are $2\text{kg}$ and $6\text{kg}$.

Write the expression for the acceleration of the horizontal acceleration of the block.

$a=\frac{F_{\text{net}}}{m}$

Here,

$m$ is the total mass of the system

$a$ is the horizontal acceleration of the system

Total mass of the system is $2\text{kg}+\text{4}\text{kg}=\text{6}\text{kg}$.

Substitute $16\text{N}$ for $F_{\text{net}}$ and $6\text{kg}$ for $m$ in the above equation to get $a$.

$\begin{array}{c}a=\frac{16\text{N}}{6\text{kg}}\\ =2.67\text{m}/{\text{s}}^{\text{2}}\end{array}$

Conclusion:

Thus, the acceleration of the system is $2.67\text{m}/{\text{s}}^{\text{2}}$.

(c)

To determine

The force acting on the $2\text{kg}$ mass by the connecting string.

Answer

The force acting on the $2\text{kg}$ mass by the connecting string is $15.6\text{N}$.

Explanation of Solution

Given info: The horizontal force acting on $2\text{kg}$ mass along the left direction is $6\text{N}$.

Let $m_1$ be the mass of $2\text{kg }$ block and $m_2$ be the mass of $4\text{kg}$ block.

Write the expression for the net force on $2\text{kg }$ mass.

$F_{\text{net1}}=m_{1}a$

Here,

$m_1$ is the mass of $2\text{kg }$ block

$F_{\text{net1}}$ is the net horizontal force on the mass $2\text{kg }$

Substitute $2\text{kg }$ for $m_1$ and $2.67\text{m}/{\text{s}}^{\text{2}}$ for $a$ in the above equation to get $F_{\text{net1}}$.

${}_{\begin{array}{c}{F}_{\text{net1}}=\left(2\text{kg }\right)\left(2.67\text{m}/{\text{s}}^{\text{2}}\right)\\ =5.34\text{N}\end{array}}$

This net force is acting along the direction of acceleration. That is, along the right direction.

Write the expression for the net force on the $2\text{kg }$ mass.

$F_{\text{net1}}=F_{\text{tension}}-F_{\text{left}}$

Here,

$F_{\text{left}}$ is the horizontal force on the left direction of mass $2\text{kg }$

$F_{\text{tension}}$ is the tension on the string

The negative sign indicate that the force of tension is along the right direction whereas the $F_{\text{left}}$ is along the left direction.

Substitute $6\text{N}$ for $F_{\text{left}}$ and $5.34\text{N}$ for $F_{\text{net1}}$ in the above equation to get $F_{\text{tension}}$.

$\begin{array}{c}5.34\text{N}=F_{\text{tension}}-6\text{N}\\ F_{\text{tension}}=11.34\text{N}\end{array}$

Therefore, the force acting on the $2\text{kg }$ by the string is equal to $11.34\text{N}$.

Conclusion:

Thus, the force acting on the $3\text{kg}$ mass by the connecting string is $11.34\text{N}$.

(d)

To determine

The net force and acceleration of $4\text{kg}$ block.

Answer

The net force on the $4\text{kg}$ block is $10.67\text{N}$ and net acceleration of the block is $2.67\text{m}/{\text{s}}^{\text{2}}$.

Explanation of Solution

Given info: The horizontal force acting on $4\text{kg}$ mass along the right direction is $30\text{N}$ and that along the left direction is $8\text{N}$

Write the expression for the net force on the $4\text{kg}$ mass.

$F_{\text{net2}}=F_{\text{right}}-\left(F_{\text{tension}}+F_{\text{left}}\right)$

Here,

$F_{\text{left}}$ is the horizontal force on the left direction of mass $4\text{kg}$

$F_{\text{right}}$ is the horizontal force on the right direction of mass $4\text{kg}$

$F_{\text{tension}}$ is the tension on the string

Substitute $30\text{N}$ for $F_{\text{right}}$ ,$11.34\text{N}$ for $F_{\text{tension}}$ and $8\text{N}$ for $F_{\text{left}}$ in the above equation to get $F_{\text{net2}}$.

$\begin{array}{c}{F}_{\text{net2}}=30\text{N}-\left(11.34\text{N}+8\text{N}\right)\\ =10.66\text{N}\end{array}$

Write the expression for the net acceleration on $4\text{kg}$ mass.

$a=\frac{F_{\text{net2}}}{m_2}$

Here,

$m_2$ is the $4\text{kg}$ mass

$a$ is the net acceleration

Substitute $4\text{kg}$ for $m_2$ and $10.66\text{N}$ for $F_{\text{net2}}$ in the above equation to get $F_{\text{net2}}$.

$\begin{array}{c}a=\frac{10.66\text{N}}{4\text{kg}}\\ =2.67\text{m}/{\text{s}}^{\text{2}}\end{array}$

This is same as the net acceleration obtained in part $b$.

Conclusion:

Thus, the net force on the $4\text{kg}$ block is $10.66\text{N}$ and net acceleration of the block is $2.67\text{m}/{\text{s}}^{\text{2}}$$3.2\text{m}/{\text{s}}^{\text{2}}$.