Chapter 4, Problem 86A

(a)

To determine

To Create:a pictorial model.

Explanation of Solution

  $\begin{array}{l}{m}_A=2.0\text{ kg}\\ m_B=2.0\text{ kg}\end{array}$

The pictorial model is given as

(b)

To determine

To Create:A physical model and with motion and free-body diagrams.

Answer to Problem 86A

A physical model and with motion and free-body diagrams are shown below.

Explanation of Solution

  $\begin{array}{l}{m}_A=2.0\text{ kg}\\ m_B=2.0\text{ kg}\end{array}$

(c)

To determine

To Find:The acceleration of smaller mass.

Answer to Problem 86A

The acceleration of smaller mass is $a=1.96{\text{ m/s}}^2$ .

Explanation of Solution

Given:

  $\begin{array}{l}{m}_A=2.0\text{ kg}\\ m_B=2.0\text{ kg}\end{array}$

Calculation:

The smaller mass moves upward direction then

  $\begin{array}{c}a=\frac{m_{B}g-m_{A}g}{m_A+m_B}\\ =\frac{(3-2)9.80}{5}\\ =1.96{\text{ m/s}}^2\end{array}$

Conclusion:

The acceleration of smaller mass is $a=1.96{\text{ m/s}}^2$ .