Two identical 0.4 kgblock (labeled 1 and 2) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper portion of Figure 9.59.

Then a constant force of 100 N to the right is applied to block 2 and at a later time the blocks are in the new positions shown in the lower portion of Figure 9.59.9.59. At this final time, the system is moving to the right and also vibrating, and the spring is stretched. (a) The following questions apply to the system modeled as a point particle. (i) What is the initial location of the point particle? (ii) How far does the point particle move? (iii) How much work was done on the particle? (iv) What is the change in translational kinetic energy of this system? (b) The following questions apply to the system modeled as an extended object. (1) How much work is done on the right-hand block? (2) How much work is done on the left-hand block? (3) What is the change of the total energy of this system? (c) Combine the results of both models to answer the following questions. (1) Assuming that the object does not get hot, what is the final value of ${\mathbf{K}}_{\mathbf{vib}}\mathbf{+}{\mathbf{U}}_{\mathbf{spring}}$for the extended system? (2) If the spring stiffness is 50 N/m, what is the final value of the vibrational kinetic energy?

The given data can be listed below as:

• The mass of the identical blocks is .$m_1=m_2=0.4\hspace{0.33em}kg$
• The initial displacement of the first block is $x_1=0$.
• The initial displacement of the second block is $x_2=0.6\hspace{0.33em}m$.
• The final displacement of the first block is $x_3=0.2\hspace{0.33em}m$.
• The final displacement of the second block is $x_4=2.0\hspace{0.33em}m$.
• The value of the applied constant force is $F=100\hspace{0.33em}N.$.
• The spring stiffness is $k=50\hspace{0.33em}\mathrm{N}/\mathrm{m}.$

The spring stiffness is described as the ability of a spring in order to resist a force. The more the stiffness, the larger is the compression of the force.

The restoring force develop in the string is given by,

$\mathit{F}\mathbf{=}\mathit{k}\mathit{\Delta }\mathit{x}\mathbf{}$

Here k is the spring constant and$\mathit{\Delta }\mathit{x}$ is the change in the length of the spring.

(i)

The equation of the initial location is expressed as:

$x=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$

Here,x is the initial location,$m_1$and $m_2$are the first and the second mass.$x_1$and$x_2$are the initial displacement of the first and the second block.

Substitute the values in the above equation.

role="math" localid="1658903437711" $$\begin{gathered} \mathrm{x}=\frac{\left(0.4\mathrm{kg}\right)\left(0\right)+\left(0.4\mathrm{kg}\right)\left(0.6\mathrm{m}\right)}{\left(0.4\mathrm{kg}\right)+\left(0.4\mathrm{kg}\right)} \\ =\frac{0+\left(0.24\mathrm{kg}.\mathrm{m}\right)}{\left(0.8\mathrm{kg}\right)} \\ =0.3\mathrm{m} \end{gathered}$$

(ii)

The equation of the displacement of the point particle is expressed as:

$y=\frac{m_1{x}_3+m_2{x}_4}{m_1+m_2}$

Here,y is the displacement of the point particle,$m_1$and $m_2$are the first and the second mass.$x_3$and$x_4$are the final displacement of the first and the second block.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{y}=\frac{\left(0.4\mathrm{kg}\right)\left(0.2\mathrm{m}\right)+\left(0.4\mathrm{kg}\right)\left(2.0\mathrm{m}\right)}{\left(0.4\mathrm{kg}\right)+\left(0.4\mathrm{kg}\right)} \\ =\frac{\left(0.08\mathrm{kg}.\mathrm{m}\right)+\left(0.8\mathrm{kg}.\mathrm{m}\right)}{\left(0.8\mathrm{kg}\right)} \\ =\frac{\left(0.88\mathrm{kg}.\mathrm{m}\right)}{\left(0.8\mathrm{kg}\right)} \\ =1.1\mathrm{m} \end{gathered}$$

(iii)

The equation of the work done is expressed as:

$w=F\left(y-x\right)$

Here,w is the work done and F is the value of the applied constant force.

Substitute the values in the above equation.

$$\begin{gathered} \mathrm{w}=\left(100\mathrm{N}\right)\left(1.1\mathrm{m}-0.3\mathrm{m}\right) \\ =\left(100\mathrm{N}\right)\left(0.8\mathrm{m}\right) \\ =80\mathrm{N}.\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}} \\ =80\mathrm{J} \end{gathered}$$

(iv)

According to the energy theorem, the work done by an object is the change in the

translational energy kinetic energy of the system. Hence, the change in translational kinetic energy of the system is 80 J.

Thus,

(i) The initial location of the point particle is 0.3 m.

(ii) The point particle move 1.1 m.

(iii) The work done on the particle is 80 J.

(iv) The change in translational kinetic energy of the system is 80 J.

(1)

The equation of the work done on the right-hand block is expressed as:

$w_1=F\left(x_4-x_2\right)$

Here,$w_1$is the work done on the right-hand block.

Substitute the values in the above equation.

role="math" localid="1658903927802" $$\begin{gathered} {\mathrm{w}}_1=\left(100\mathrm{N}\right)\left(2.0\mathrm{m}-0.6\mathrm{m}\right) \\ =\left(100\mathrm{N}\right)\left(1.4\mathrm{m}\right) \\ =140\mathrm{N}.\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}} \\ =140\mathrm{J} \end{gathered}$$

(2)

The equation of the work done on the left-hand block is expressed as:

$w_2=F\left(x_3-x_1\right)$

Here,$w_2$is the work done on the left-hand block.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{w}}_2=\left(100\mathrm{N}\right)\left(0.2\mathrm{m}-0\mathrm{m}\right) \\ =\left(100\mathrm{N}\right)\left(0.2\mathrm{m}\right) \\ =20\mathrm{N}.\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}} \\ =20\mathrm{J} \end{gathered}$$

(3)

According to the energy theorem, the total work done by an object is the change in the total energy of the system. Hence, the change in total energy of the system is $140\hspace{0.33em}\mathrm{J}+20\mathrm{J}=160\hspace{0.33em}\mathrm{J}.$

Thus,

(1) The work done on the right-hand block is 140 J.

(2) The work done on the left-hand block is 20 J.

(3) The change in the total energy of the system is 160 J.

(1)

If the object does not get hot, then the final value of$K_{vib}+U_{Spring}$will be the total energy of the system which is 160 J.

(2)

The equation of the total energy is expressed as:

$E=K_{vib}+U_{Spring}$

Here,$U_{Spring}$is the potential energy of the spring,$k_{vib}$ is the vibrational kinetic energy andE is the total energy.

The total energy is the change in the total energy of the system.

The equation of the potential energy of the spring can be expressed as:

$U_{Spring}=\frac{1}{2}k\left(x_4-x_2\right)$

Here,k is the spring constant.

From the above equation, the equation of the vibrational kinetic energy is expressed as:

$K_{vib}=E-\frac{1}{2}k{\left[\left(x_4-x_2\right)-\left(x_3-x_1\right)\right]}^2$

Here,$k_{vib}$ is the vibrational kinetic energy andk is the spring constant.

Substitute the values in the above equation.

$$\begin{gathered} k_{vib}=160\mathrm{J}-\frac{1}{2}\left(50\mathrm{N}/\mathrm{m}\right){\left(1.4\mathrm{m}-0.2\mathrm{m}\right)}^2 \\ =160\mathrm{J}-\left(25\mathrm{N}/\mathrm{m}\right)\left(1.4{\mathrm{m}}^2\right) \\ =160\mathrm{J}-\left(35\mathrm{N}.\mathrm{m}\times \frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}\right) \\ =125\mathrm{J} \end{gathered}$$

Thus,

(1) The final value of$K_{vib}+U_{Spring}$ is 160 J.

(2) The final value of the vibrational kinetic energy is 125 J.