Question

A box contains machinery that can rotate. The total mass of the box plus the machinery is$\mathbf{7}\mathbf{}\mathbf{kg}$. A string wound around the machinery comes out through a small hole in the top of the box. Initially the box sits on the ground, and the machinery inside is not rotating (left side of Figure 9.61). Then you pull upward on the string with a force of constant magnitude . At an instant when you have pulled 0.6mof string out of the box (indicated on the right side of Figure 9.61), the box has risen a distance of 0.2 mand the machinery inside is rotating.

POINT PARTICLE SYSTEM (a) List all the forms of energy that change for the point particle system during this process. (b) What is the$y$component of the displacement of the point particle system during this process? (c) What is the ycomponent of the net force acting on the point particle system during this process? (d) What is the distance through which the net force acts on the point particle system? (e) How much work is done on the point particle system during this process? (f) What is the speed of the box at the instant shown in the right side of Figure 9.61? (g) Why is it not possible to find the rotational kinetic energy of the machinery inside the box by considering only the point particle system?

EXTENDED SYSTEM (h) the extended system consists of the box, the machinery inside the box, and the string. List all the forms of energy that change for the extended system during this process. (i) What is the translational kinetic energy of the extended system, at the instant shown in the right side of Figure 9.61? (j) What is the distance through which the gravitational force acts on the extended system? (k) How much work is done on the system by the gravitational force? (I) what is the distance through which your hand moves? (m) How much work do you do on the extended system? (n) At the instant shown in the right side of Figure 9.61, what is the total kinetic energy of the extended system? (o) what is the rotational kinetic energy of the machinery inside the box?

Short answer

POINT PARTICLE SYSTEM

(a) The translational kinetic energy changes for the point particle system during this process.

(b)Thecomponent of the displacement of the point particle system during the process is$0.6\mathrm{m}$.

(c) The component of the net force acting on the point particle system during this process is $61.4\mathrm{N}$.

(d)The distance through which the net force acts on the point particle system is 0.6m.

(e) The work done on the point particle system during this process is 36.84 J.

(f) The speed of the box at the instant is 3.244 m/s.

(g) As the point particle does not have rotational kinetic energy, that is the main reason for not finding the rotational kinetic energy of the machinery inside the box by considering only the point particle system.

EXTENDED SYSTEM

(h) The forms of energy that change for the extended system during this process are translational and rotational kinetic energy.

(i) The transitional kinetic energy of the extended system is 36.83 J.

(j) The distance the gravitational force acts on the extended system are 0.6m.

(k) The work done on the system by the gravitational force is41.16 J .

(l) The distance through which the hand moves is 0.8m.

(m) Work done by the extended system is 104 J.

(n) The total kinetic energy of the extended system is 62.84 J.

(o) The rotational kinetic energy of the machinery inside the box is 26.01 J.

Step-by-step solution

Identification of given data

• The mass of the box plus the machinery is, m=7 kg
• The force needed to pull the string is, F=130N
• The distance to which the box rose is, h =0.2m
• The distance by which the string has been pulled out of the box, d=0.6m

 Step 2: Definition of work done and kinetic energy

Work done is the displacement caused by the object by applying force. The general expression for the work done is as follows,

W=Fd

Here,F is the force applied and is the distance at which the force is applied.

Kinetic energy is motion energy which is caused by the object during motion.

(a) Description of the change forms of energy

When any object is fixed at a point, it will be pulled with some force, and then the energies related to that object are being changed is the translational kinetic energy.

Thus, the translational kinetic energy changes for the point particle system during this process.

(b) Calculation for the component for the displacement

About the pointed system and the comprehensive system, only one displacement type occurs the distance by which the string is pulled out of the box. Hence, the y component is the distance the string is pulled out of the box.

Thus, the y component of the displacement of the point particle system during the process is 0.6m.

(c) Calculation for the component of the net force

The equation of the net force of the y component is expressed as:

$F_y=F-mg$

Here, F is the force needed to pull the string, m is the mass of the box plus the machinery and g is the acceleration due to gravity that has the value $9.8m/s^2$.

Substitute the values in the above equation.

$$\begin{gathered} F_y=130\mathrm{N}-7\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2 \\ =130\mathrm{N}-68.6\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2 \\ =130\mathrm{N}-68.6\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\times \left(\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =61.4\mathrm{N} \end{gathered}$$

Thus, the component of the net force acting on the point particle system during this process is 61.4N

(d) Calculation for the distance of the net force

The distance through which net force acts is the distance by which the extended system is raised by a pointed system that is the distance the string is pulled out of the box.

Thus, the distance the net force acts on the point particle system is 0.6m.

(e) Calculation for the total work done on the pointed particle

The equation of the total work done is expressed as:

$w=F_y.h$

Here, $F_y$is the force of the y component and is the distance through which the string has pulled out of the box.

Substitute the values in the above equation

$$\begin{gathered} w=61.4\mathrm{N}\times 0.6\mathrm{m} \\ =36.84\mathrm{N}.\mathrm{m} \\ =36.84\mathrm{N}.\mathrm{m}\times \left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ =36.84\mathrm{J} \end{gathered}$$

Thus, the work done on the point particle system during this process is 36.84 J.

(f) Calculation for the speed of the box

The equation of the speed of the box is expressed as:

$$\begin{gathered} w=\frac{1}{2}m{v}^2 \\ v=\sqrt{\frac{2w}{m}} \end{gathered}$$

Here, w is the work done which is the same as that of the kinetic energy, m is the mass of the box plus the machinery and v is the speed of the box.

Substitute the values in the above equation.

$$\begin{gathered} v=\sqrt{\frac{2\times 3684\mathrm{J}}{7\mathrm{kg}}} \\ =\sqrt{10.525\mathrm{J}/\mathrm{kg}} \\ =\sqrt{10.525\frac{\mathrm{J}}{\mathrm{kg}}\times \left(\frac{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}{1\mathrm{J}}\right)} \\ =\sqrt{10.525\frac{\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}{\mathrm{kg}}} \\ =3.244\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the speed of the box at the instant is 3.244 m/s.

 Step 9: (g) Calculation for the rotational kinetic energy of the machinery

The rotational kinetic energy of the machinery inside the box is not calculated because no energy principle can apply to the rotating objects. However, a point particle does not contain rotational kinetic energy.

Thus, as the point particle does not have rotational kinetic energy, that is the main reason for not finding the rotational kinetic energy of the machinery inside the box by considering only the point particle system.

(h) Description of the change forms of energy

When any object is fixed at a point, and after that, it will be pulled with some amount of force, then the energies related to that object are changed is as follows:

i) Translational kinetic energy

ii) Rotational kinetic energy

Thus, the forms of energy that change for the extended system during this process are translational and rotational kinetic energy.

(i) Calculation for transitional kinetic energy

The equation of the translational kinetic energy can be expressed as:

$k_{tras}=\frac{1}{2}m{v}^2$

Here,m is the mass of the box plus the machinery and v is the speed of the box at the instant.

Substitute the values in the above equation.

$$\begin{gathered} k_{tras}=\frac{1}{2}\times 7\mathrm{kg}\times \left(3.244\mathrm{m}/\mathrm{s}\right) \\ =36.83{\mathrm{kgm}}^2/{\mathrm{s}}^2\left(\frac{1\mathrm{J}}{1{\mathrm{kgm}}^2/{\mathrm{s}}^2}\right) \\ =36.83\mathrm{J} \end{gathered}$$

Thus, the transitional kinetic energy of the extended system is $36.83\mathrm{J}$.

(j) Determination of the distance the gravitational force acts on the extended system

The distance through which the gravitational force mainly acts on the extended

The system is the distance that the string has pulled out from the box, that is 6.0 m .

Thus, the distance the gravitational force acts on the extended system are 6.0 m .

(k) Calculation for work done by gravitational force

The equation of the work done is expressed as:

$w_g=mg{h}_1$

Here, m is the mass of the box plus the machinery, g is the acceleration due to gravity $h_1$and is the distance the gravitational force acts on the extended system.

Substitute the values in the above equation.

$$\begin{gathered} w_g=7\mathrm{kg}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times 0.6\mathrm{m} \\ =41.16\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\times \frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2{\mathrm{s}}^2} \\ =41.16\mathrm{J} \end{gathered}$$

Thus, the work done on the system by the gravitational force is $41.16\mathrm{J}$.

(l) Determination of the distance through which the hand moves

The equation of the distance through which the hand moves is calculated as:

$a=h+d$

Here, h is the distance to which the box rose and d is the distance through which the string was pulled out from the box.

Substitute the values in the above equation.

$$\begin{gathered} a=\left(0.2\mathrm{m}+0.6\mathrm{m}\right) \\ =0.8\mathrm{m} \end{gathered}$$

Thus, the distance through which the hand moves is 0.8 m.

(m) Calculation for work done by the extended system

The equation of the work done by the extended system is calculated as:

$w_e=F\left(h+d\right)$

Here, F is the force needed to pull the string, h is the distance to which the box rose and d is the distance through which the string is pulled out from the box.

Substitute the values in the above equation.

$$\begin{gathered} w_e=130\mathrm{N}\times \left(0.2\mathrm{m}+0.6\mathrm{m}\right) \\ =104\mathrm{N}.\mathrm{m} \\ =104\mathrm{N}.\mathrm{m}\times \left(\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \end{gathered}$$

Thus, work done by the extended system is 104 J.

(n) Calculation for total kinetic energy

The equation for calculating the total kinetic energy is expressed as:

$k_t=w_e-w_g$

Here,$w_e$is the work done by the extended system and$w_g$ is the work done by the gravitational force.

Substitute the values in the above equation.

$$\begin{gathered} k_t=104\mathrm{J}-41.16\mathrm{J} \\ =62.84\mathrm{J} \end{gathered}$$

Thus, the total kinetic energy of the extended system is 62.84 J.

(o) Calculation for rotational kinetic energy

The equation of the rotational kinetic energy is expressed as:

$k_{rot}=k_{total}-k_{tras}$

Here,$k_{total}$is the total kinetic energy and$k_{tras}$ is the translational kinetic energy.

Substitute the values in the above equation.

$$\begin{gathered} k_{rot}=62.84\mathrm{J}-36.83\mathrm{J} \\ =26.01\mathrm{J} \end{gathered}$$

Thus, the rotational kinetic energy of the machinery inside the box is 26.01 J