Chapter 7, Problem 7.13OQ

(i)

To determine

The rank of following gravitational acceleration for the following Falling object.

Answer to Problem 7.13OQ

The rank of the following gravitational acceleration for the falling object $g_1=g_2=g_3=g_4$.

Explanation of Solution

Given info: A $2\text{kg}$ object falling from the $5\text{cm}$ above the floor, a $2\text{kg}$ object falling from the $120\text{cm}$ above the floor, a $3\text{kg}$ object falling from the $120\text{cm}$ above the floor and a $3\text{kg}$ object falling from the $80\text{cm}$ above the floor.

Write the expression for the gravitational acceleration at the height $h$ from the surface of the earth.

$g^{\prime }=g{\left(1+\frac{h}{R}\right)}^{-2}$

Here,

$R$ is the radius of the earth.

$g$ is the gravitational acceleration.

The radius of the earth is much greater than the height from the point where the object is falling so that the change in the gravitation acceleration is negligible. The value of the gravitational acceleration for all object are equal to $9.81\text{m}/{\text{s}}^{\text{2}}$.

The rank of the following gravitational acceleration for the falling object $g_1,g_2,g_3,g_4$.

$\begin{array}{c}{g}_1=g_2\\ g_2=g_3\\ g_3=g_4\\ g_4=9.81\text{m}/{\text{s}}^{\text{2}}\end{array}$

Here,

$g_1$ is the gravitation acceleration for the $2\text{kg}$ object falling from the height $5\text{cm}$ above the floor.

$g_2$ is the gravitation acceleration for the $2\text{kg}$ object falling from the height $120\text{cm}$ above the floor.

$g_3$ is the gravitation acceleration for the $3\text{kg}$ object falling from the height $120\text{cm}$ above the floor.

$g_4$ is the gravitation acceleration for the $3\text{kg}$ object falling from the height $80\text{cm}$ above the floor.

Conclusion:

Therefore, the rank of following gravitational acceleration for the falling object $g_1=g_2=g_3=g_4$.

(ii)

To determine

The rank of following gravitational forces for the following Falling object.

Answer to Problem 7.13OQ

The rank of the following gravitational forces for the falling object is $F_3=F_4>F_1=F_2$.

Explanation of Solution

Given info: A $2\text{kg}$ object falling from the $5\text{cm}$ above the floor, a $2\text{kg}$ object falling from the $120\text{cm}$ above the floor, a $3\text{kg}$ object falling from the $120\text{cm}$ above the floor and a $3\text{kg}$ object falling from the $80\text{cm}$ above the floor.

Write the expression for the gravitational force.

$F=mg$ (1)

Here,

$g$ is the gravitational acceleration of earth.

For object of mass $2\text{kg}$ and distance $5\text{cm}$:

Substitute $2\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above expression for the force acting on the object falling from $5\text{cm}$.

$\begin{array}{c}{F}_1=\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =19.62\text{N}\end{array}$

Thus, the force on the $2\text{kg}$ mass fall from $5\text{cm}$ is $19.62\text{N}$.

For object of mass $2\text{kg}$ and distance $120\text{cm}$:

Substitute $2\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the expression (1) for the force acting on the object falling from $120\text{cm}$.

$\begin{array}{c}{F}_2=\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =19.62\text{N}\end{array}$

Thus, the force on the $2\text{kg}$ mass fall from $120\text{cm}$ is $19.62\text{N}$.

For object of mass $3\text{kg}$ and distance $120\text{cm}$:

Substitute $3\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the expression (1) for the force acting on the object falling from $120\text{cm}$.

$\begin{array}{c}{F}_3=\left(3\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =29.43\text{N}\end{array}$

Thus, the force on the $3\text{kg}$ mass fall from $120\text{cm}$ is $29.43\text{N}$.

For object of mass $3\text{kg}$ and distance $80\text{cm}$:

Substitute $3\text{kg}$ for $m$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the expression (1) for the force acting on the object falling from $80\text{cm}$.

$\begin{array}{c}{F}_4=\left(3\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\\ =29.43\text{N}\end{array}$

Thus, the force on the $3\text{kg}$ mass fall from $80\text{cm}$ is $29.43\text{N}$.

From the value of the forces the ranking of the following gravitational forces for the falling object.

$F_3=F_4>F_1=F_2$

Conclusion:

Therefore, the rank of the following gravitational forces for the falling object is $F_3=F_4>F_1=F_2$.

(iii)

To determine

The rank of following gravitational potential energy for the following Falling object.

Answer to Problem 7.13OQ

The rank of the following gravitational potential energy for the falling object is $U_3>U_2=U_4>U_1$.

Explanation of Solution

Given info: A $2\text{kg}$ object falling from the $5\text{cm}$ above the floor, a $2\text{kg}$ object falling from the $120\text{cm}$ above the floor, a $3\text{kg}$ object falling from the $120\text{cm}$ above the floor and a $3\text{kg}$ object falling from the $80\text{cm}$ above the floor.

Write the expression for the gravitational potential energy for the falling of object.

$U=mgh$ (1)

Here,

$g$ is the gravitational acceleration of earth.

$m$ is the mass of the falling object.

$h$ is the height of the object from the floor.

For object of mass $2\text{kg}$ and distance $5\text{cm}$:

Substitute $2\text{kg}$ for $m$, $5\text{cm}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above expression for the potential energy on the object falling from $5\text{cm}$.

$\begin{array}{c}{U}_1=\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(5\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =0.981\text{J}\end{array}$

Thus, the potential energy for the $2\text{kg}$ mass falling from $5\text{cm}$ is $0.981\text{J}$.

For object of mass $2\text{kg}$ and distance $120\text{cm}$:

Substitute $2\text{kg}$ for $m$, $120\text{cm}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the equation (1) for the potential energy on the object falling from $120\text{cm}$.

$\begin{array}{c}{U}_2=\left(2\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(120\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =23.54\text{J}\end{array}$

Thus, the potential energy for the $2\text{kg}$ mass falling from $120\text{cm}$ is $23.54\text{J}$.

For object of mass $3\text{kg}$ and distance $120\text{cm}$:

Substitute $3\text{kg}$ for $m$, $120\text{cm}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the equation (1) for the potential energy on the object falling from $120\text{cm}$.

$\begin{array}{c}{U}_3=\left(3\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(120\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =35.316\text{J}\end{array}$

Thus, the potential energy for the $3\text{kg}$ mass falling from $120\text{cm}$ is $35.316\text{J}$.

For object of mass $3\text{kg}$ and distance $80\text{cm}$:

Substitute $3\text{kg}$ for $m$, $80\text{cm}$ for $h$ and $9.81\text{m}/{\text{s}}^{\text{2}}$ for $g$ in the above expression for the potential energy on the object falling from $80\text{cm}$.

$\begin{array}{c}{U}_4=\left(3\text{kg}\right)\left(9.81\text{m}/{\text{s}}^{\text{2}}\right)\left(80\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =23.54\text{J}\end{array}$

Thus, the potential energy for the $3\text{kg}$ mass falling from $80\text{cm}$ is $23.54\text{J}$.

From the value of the energies the ranking of the following gravitational potential energies for the falling object.

$U_3>U_2=U_4>U_1$

Conclusion:

Therefore, the rank of the following gravitational potential energy for the falling object is $U_3>U_2=U_4>U_1$.