Chapter 4, Problem 82A

(a)

To determine

Theacceleration of the two blocks $\text{A}$ and $\text{B}$ .

Answer to Problem 82A

The acceleration of the blocks is $2.3{\text{m/s}}^2$ towards right.

Explanation of Solution

Given:

The arrangements of the blocks are shown.

Consider two blocks $\text{A}$ and $\text{B}$ are attached with each other as shown in the arrangement.

Consider the mass of the block $\text{A}$ is $m_{\text{A}}=4.3\text{kg}$ .

Consider the mass of the block $\text{B}$ is $m_{\text{B}}=5.4\text{kg}$ .

Consider the block $\text{A}$ is pushed by a force $F=22.5\text{N}$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Here, $m$ is the mass and $a$ is the acceleration of the object.

Calculation:

Consider both the blocks $\text{A}$ and $\text{B}$ acts as a system.

Consider the entire assembly is pushed by a force of $F=22.5\text{N}$ towards the right.

Take the direction towards right as positive.

Find the acceleration of the blocks as follows:

Apply Newton’s Second Law of Motion,

  $\begin{array}{c}a=\frac{F}{m_{\text{A}}+m_{\text{B}}}\\ a=\frac{22.5\text{N}\times \left(\frac{1\text{kg}\cdot {\text{m/s}}^2}{1\text{N}}\right)}{4.3\text{kg}+5.4\text{kg}}\\ a=\frac{22.5\text{kg}\cdot {\text{m/s}}^2}{9.7\text{kg}}\\ a=2.3{\text{m/s}}^2\left(\text{towards the right}\right)\end{array}$

Conclusion:

Thus, the acceleration of the blocks is $2.32{\text{m/s}}^2$ towards the right.

(b)

To determine

The force exerted by the block of mass $m_{\text{A}}=4.3\text{kg}$ on the block of mass $m_{\text{B}}=5.4\text{kg}$ .

Answer to Problem 82A

The force exerted by the block of mass $4.3\text{kg}$ on the block of mass $5.4\text{kg}$ is $12\text{N}$ towards right.

Explanation of Solution

Given:

The arrangements of the blocks are shown.

Consider two blocks $\text{A}$ and $\text{B}$ are attached with each other as shown in the arrangement.

Consider the mass of the block $\text{A}$ is $m_{\text{A}}=4.3\text{kg}$ .

Consider the mass of the block $\text{B}$ is $m_{\text{B}}=5.4\text{kg}$ .

Formula used:

Newton’s Second Law:

The object’s acceleration is equivalent to the sum of the forces acting on the object divided by the object’s mass. That is,

  $\begin{array}{l}a=\frac{F_{\text{net}}}{m}\\ F_{\text{net}}=ma\end{array}$

Here, $m$ is the mass and $a$ is the acceleration of the object.

Calculation:

Consider both the block of mass $m_{\text{B}}=5.4\text{kg}$ acts as a system.

Take the direction towards right as positive.

Consider the force exerted by the block of mass $m_{\text{A}}=4.3\text{kg}$ on $m_{\text{B}}=5.4\text{kg}$ is denoted by $F_{\text{AB}}$ .

The acceleration of the blocks is $a=2.3{\text{m/s}}^2$ towards the right.

Find the force exerted by the block $\text{A}$ on block $\text{B}$ as follows:

Apply Newton’s Second Law of Motion,

  $\begin{array}{c}{F}_{\text{AB}}=m_{\text{B}}a\\ F_{\text{AB}}=5.4\text{kg}\times 2.3{\text{m/s}}^2\\ F_{\text{AB}}=12\text{kg}\cdot {\text{m/s}}^2\times \left(\frac{1\text{N}}{1\text{kg}\cdot {\text{m/s}}^2}\right)\\ F_{\text{AB}}=12\text{N}\left(\text{towards the right}\right)\end{array}$

Conclusion:

Thus, the force exerted by the block of mass $4.3\text{kg}$ on the block of mass $5.4\text{kg}$ is 12 N towards right.

(c)

To determine

The force exerted by the block of mass $m_{\text{B}}=5.4\text{kg}$ on the block of mass $m_{\text{A}}=4.3\text{kg}$ .

Answer to Problem 82A

The force exerted by the block of mass $5.4\text{kg}$ on the block of mass $4.3\text{kg}$ is 12 N towards left.

Explanation of Solution

Given info:

The arrangements of the blocks are shown.

Consider two blocks $\text{A}$ and $\text{B}$ are attached with each other as shown in the arrangement.

Consider the mass of the block $\text{A}$ is $m_{\text{A}}=4.3\text{kg}$ .

Consider the mass of the block $\text{B}$ is $m_{\text{B}}=5.4\text{kg}$ .

Formula used:

Newton’s Third Law of Motion:

According to Newton’s Third Law, for every action there is an equal and opposite reaction. When a body $A$ applies a force on body $B$ then the body $B$ exerts an equal and opposite force on body $A$ .

  $F_{\text{A on B}}=-F_{\text{B on A}}$

Calculation:

Refer Part (b).

The force exerted by the block of mass $4.3\text{kg}$ on the block of mass $5.4\text{kg}$ is 12 N towards right.

Consider the force exerted by the block of mass $5.4\text{kg}$ on the block of mass $4.3\text{kg}$ is denoted by $F_{\text{BA}}$ .

Apply Newton’s Third Law of Motion,

The block of mass $5.4\text{kg}$ also exerts an equal and opposite force on the block of mass $4.3\text{kg}$ .

  $\begin{array}{l}{F}_{\text{BA}}=-F_{\text{AB}}\\ F_{\text{BA}}=-12\text{N}\end{array}$

The negative sign indicates the direction of the force is towards the left.

Conclusion:

Thus, the force exerted by the block of mass $5.4\text{kg}$ on the block of mass $4.3\text{kg}$ is 12 N towards left.