Chapter 4, Problem 31P

To determine

Part (a) To Determine:

Acceleration of each block.

Answer to Problem 31P

Solution:

Acceleration of each block (a)= 2.72 m/s².

Explanation of Solution

Let a be the acceleration of the system, g be the acceleration due to gravity, T be the tension in the string.

For m_A:

$T=m_{A}a$.

For m_B:

$m_{B}g-T=m_{B}a$.

Adding to eliminate T:

$m_{B}g=\left(m_A+m_B\right)a$…………(1)

$a=\left(m_{B}g\right)/\left(m_A+m_B\right)$.

$=\left(5\times 9.81\right)/\left(13+5\right)$.

$=2.72{\text{ m/s}}^{\text{2}}$.

Given:

$m_A=13.0\text{ kg}$ And $m_B=5.0\text{ kg}$.

Formula used:

$T=m_{A}a$ And $m_{B}g-T=m_{B}a$.

Calculation:

$a=\left(m_{B}g\right)/\left(m_A+m_B\right)$.

$=\left(5\times 9.81\right)/\left(13+5\right)$.

$=2.72{\text{ m/s}}^{\text{2}}$.

To determine

Part (b) To Determine:

How long does it take to reach the edge of the table if the system can move freely?

Answer to Problem 31P

Solution:

$Timetaken\left(t\right)=0.919\text{ s}$.

Explanation of Solution

Let:

S be the distance to the edge of the table,

t be the time taken.

$S=a{t}^2/2$.

$t=\sqrt{\frac{2s}{a}}$.

$t=\sqrt{\frac{2\times 1.150}{2.72}}$.

So, time to reach the edge of the table if the system can move freely is = 0.919 s.

Given:

Masses $m_A=13.0\text{ kg}$ and $m_B=5.0\text{ kg}$. Initially, m_A is at rest 1.250 m from the edge of the table.

Formula used:

$S=a{t}^2/2$.

Calculation:

$t=\sqrt{\frac{2\times 1.150}{2.72}}$.

$Timetaken\left(t\right)=0.919\text{ s}$.

To determine

Part (c) To Determine:

Mass of block A

Answer to Problem 31P

Solution:

$Valueofmassofblock{m}_A=99\text{kg}$.

Explanation of Solution

Given:

m_A=13.0 kg and m_B= 5.0 kg in Fig. 4-53. If m_Bis changed to m_B =1.0 kg, and acceleration of the system is to be kept at 1/100 g.

Formula used:

For m_A:

$T=m_{A}a$

For m_B:

$m_{B}g=\left(m_A+m_B\right)a$.

Calculation:

Let:

a be the acceleration of the system, g be the acceleration due to gravity, T be the tension in the string.

For m_A:

$T=m_{A}a$.

For m_B:

Adding to eliminate T:

$m_{B}g=\left(m_A+m_B\right)a$…………(1)

From (1):

$\left(m_A+m_B\right)\left(a/g\right)=m_B$.

$m_A\left(a/g\right)=m_B\left(1-a/g\right)$.

$m_A=m_B\left(g/a-1\right)$

$=1\left(100-1\right)$.

$=99\text{kg}$.