Chapter 8, Problem 26AP

Review. As shown in Figure P8.26, a light string that does not stretch changes from horizontal to vertical as it passes over the edge of a table. The string connects m₁, a 3.50-kg block originally at rest on the horizontal table at a height h = 1.20 m above the floor, to m₂, a hanging 1.90-kg block originally a distance d = 0.900 m above the floor. Neither the surface of the table nor its edge exerts a force of kinetic friction. The blocks start to move from rest. The sliding block m₁ is projected horizontally after reaching the edge of the table. The hanging block m₂ stops without bouncing when it strikes the floor. Consider the two blocks plus the Earth as the system. (a) Find the speed at which m₁ leaves the edge of the table. (b) Find the impact speed of m₁ on the floor. (c) What is the shortest length of the string so that it does not go taut while m₁ is in flight? (d) Is the energy of the system when it is released from rest equal to the energy of the system just before m₁ strikes the ground? (e) Why or why not?

Figure P8.26

To determine

To determine: The speed at which block of mass $m_1$ leaves the edge of the table.

Answer to Problem 26AP

Answer: The speed at which block of mass $m_1$ leaves the edge of the table is $\text{2}\text{.02}\text{m}/\text{s}$.

Explanation of Solution

Given Info: A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass $3.5\text{kg}$ block originally at rest on the horizontal table at the height $1.20\text{m}$ above the floor to hanging the mass $1.9\text{kg}$ block originally a distance $0.9\text{m}$ above the floor.

Explanation:

The free body diagram of the body of mass $m_1$ and mass $m_2$.

Figure I

Formula to calculate acceleration of the block by D’Alembert’s principle for mass $m_1$ is,

$T=m_{1}a$

• $T$ is the tension in the string.
• $m_1$ is the mass of block $1$.
• $a$ is the acceleration of the blocks.

Formula to calculate acceleration of the block by D’Alembert’s principle for mass $m_2$ is,

$T-m_{2}g=m_{2}a$                                       (I)

• $m_2$ is the mass of block $2$.
• $g$ is the acceleration due to gravity.

Substitute $m_{1}a$ for $T$ in equation (I).

$\begin{array}{c}{m}_{1}a-m_{2}g=m_{2}a\\ a=\frac{m_{2}g}{m_2+m_1}\end{array}$

Formula to calculate the distance covered by the block $m_2$ is,

$d=ut+\frac{1}{2}a{t}^2$

• $d$ is the distance covered by the covered by the block $m_2$.
• $u$ is the initial velocity of the block of mass $m_2$.
• $t$ is the time taken by block of mass $m_2$ to hit the floor.
• $a$ is the acceleration of the block of mass $m_2$.

Substitute $\frac{m_{2}g}{m_2+m_1}$ for $a$ to find $s$.

$d=ut+\frac{1}{2}\times \frac{m_{2}g}{m_2+m_1}\times t^2$

Substitute $3.5\text{kg}$ for $m_1$, $0$ for $u$, $1.9\text{kg}$ for $m_2$, $0.9\text{m}$ for $d$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $t$.

$\begin{array}{c}0.9\text{m}=0\times t+\frac{1}{2}\times \frac{1.9\text{kg}\times 9.81\text{m}/{\text{s}}^{\text{2}}}{1.9\text{kg}+3.5\text{kg}}\times t^2\\ t=0.722\text{s}\end{array}$

Formula to calculate the speed of block of mass $m_1$ is,

$v=u+at$

• $v$ is the final velocity of block $1$.

Substitute $\frac{m_{2}g}{m_2+m_1}$ for $a$ to find $v$.

$v=u+\frac{m_{2}g}{m_2+m_1}t$

Substitute $0.722\text{s}$ for $t$, $0$ for $u$, $3.5\text{kg}$ for $m_1$, $1.9\text{kg}$ for $m_2$, and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $v$.

$\begin{array}{c}v=0+\frac{1.9\text{kg}\times 9.81\text{m}/{\text{s}}^{\text{2}}}{1.9\text{kg}+3.5\text{kg}}\times 0.722\text{s}\\ =\text{2}\text{.02}\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the speed at which block of mass $m_1$ leaves the edge of the table is $\text{2}\text{.02}\text{m}/\text{s}$.

To determine

To determine: The impact speed of the block of mass $m_1$ on the floor.

Answer to Problem 26AP

Answer: The impact speed of the block of mass $m_1$ on the floor is $5.25\text{m}/\text{s}$.

Explanation of Solution

Given info: A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass $3.5\text{kg}$ block originally at rest on the horizontal table at the height $1.20\text{m}$ above the floor to hanging the mass $1.9\text{kg}$ block originally a distance $0.9\text{m}$ above the floor.

Explanation:

Formula to calculate the impact speed of the block of mass $m_1$ on the floor is,

$V=\sqrt{v^2+2gh}$

• $V$ is the impact speed of the block of mass $m_1$.
• $v$ is the speed of the block of mass $m_1$.
• $h$ is the distance between block of mass $m_1$ and the floo.

Substitute $\text{2}\text{.02}\text{m}/\text{s}$ for $v$, $1.2\text{m}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $V$.

$\begin{array}{c}V=\sqrt{{\left(\text{2}\text{.02}\text{m}/\text{s}\right)}^2+2\times 9.81\text{m}/{\text{s}}^{\text{2}}\times 1.2\text{m}}\\ =5.25\text{m}/\text{s}\end{array}$

Conclusion:

Therefore, the impact speed of the block of mass $m_1$ on the floor is $5.25\text{m}/\text{s}$.

To determine

To determine: The shortest length of the string so that it does not go taut while $m_1$ is in flight

Answer to Problem 26AP

Answer: The shortest length of the string so that it does not go taut while $m_1$ is in flight is $1\text{m}$.

Explanation of Solution

Explanation:

Given information:

A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass $3.5\text{kg}$ block originally at rest on the horizontal table at the height $1.20\text{m}$ above the floor to hanging the mass $1.9\text{kg}$ block originally a distance $0.9\text{m}$ above the floor.

Formula to calculate the shortest distance of the string is,

$s=v\sqrt{\frac{2h}{g}}$

• $s$ is the shortest length of the string so that it does not go taut while $m_1$ is in flight.

Substitute $\text{2}\text{.02}\text{m}/\text{s}$ for $v$, $1.2\text{m}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ to find $s$.

$\begin{array}{c}s=\text{2}\text{.02}\text{m}/\text{s}\sqrt{\frac{2\times 1.2\text{m}}{9.81\text{m}/{\text{s}}^{\text{2}}}}\\ =1\text{m}\end{array}$

Conclusion:

Therefore, the shortest length of the string so that it does not go taut while $m_1$ is in flight is $1\text{m}$.

To determine

To determine: Conservation of energy states that the energy of any isolated system remains constant as the energy can neither be created nor destroyed.

Answer to Problem 26AP

Therefore, the energy of the system when it is released from rest equal to the system just before $m_1$ strikes the ground as there is no frictional force present.

Explanation of Solution

Given info:

Explanation: A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass $3.5\text{kg}$ block originally at rest on the horizontal table at the height $1.20\text{m}$ above the floor to hanging the mass $1.9\text{kg}$ block originally a distance $0.9\text{m}$ above the floor.

There is no frictional force present here so, any type of losses will not occur here. Thus, the energy of the system when it is released from rest equal to the system just before $m_1$ strikes the ground.

Conclusion:

Therefore, the energy of the system when it is released from rest equal to the system just before $m_1$ strikes the ground as there is no frictional force present.

To determine

To determine: The energy of the system when it is released from rest equal to the system just before $m_1$ why or why not.

Answer to Problem 26AP

Therefore, the energy is conserved in the absence of frictional force so, energy of the system when it is released from rest equal to the system just before $m_1$ strikes the ground.

Explanation of Solution

Given info:

Explanation: Conservation of energy is state that the energy of any isolated system is remains constant. So the energy can neither be created nor destroyed.

A light string that does not stretch changes from horizontal to vertical its passes over the edge of the table. The string connects of mass $3.5\text{kg}$ block originally at rest on the horizontal table at the height $1.20\text{m}$ above the floor to hanging the mass $1.9\text{kg}$ block originally a distance $0.9\text{m}$ above the floor.

There is no frictional force present here so, any type of losses will not occur here. Thus, the energy of the system when it is released from rest equal to the system just before $m_1$ strikes the ground. So, energy is always conserved in ordinary conditions.

Conclusion:

Therefore, the energy is conserved in the absence of frictional force so, energy of the system when it is released from rest equal to the system just before $m_1$ strikes the ground.