91Ó°ÊÓ

The given data can be listed below as,

The change in the energy is described as the addition of the work done by the system and the heat added to the system. Moreover, the change in the energy also states that the energy can also be converted from one form to another form.

The free body diagram showing the acting forces along with the distance has been provided below-

Here, the force exerted on the first block whose height is $0.3\mathrm{m}\mathrm{is}2Mg$and the force exerted on the second block whose height is$1.2\mathrm{m}\mathrm{is}\mathrm{more}\mathrm{than}2Mg$.

The equation of the energy of the point particle system can be expressed as:

$∆E=W+Q$

Here, $W$is the work done by the system and$Q$is the heat added to the system.

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

$∆K_{\mathrm{translation}}=F_{netexternal}∆{\stackrel{→}{r}}_{CM}$

Here,$F_{netextenal}$is the net external force and $∆{\stackrel{→}{r}}_{CM}$is the difference between the initial and the final center of mass

The equation of the net force can be expressed as:

$F_{net}=F-F_{grav}$

…(¾±¾±)

Here, $F$is the force exerted and $F_{grav}$is the gravitational force.

The equation of the difference of the final and the initial center of mass can be expressed as:

$\left(y_{CM.f}-Y_{CM.I}\right)=\left(\frac{y_{1f}+y_{1f}}{2}-\frac{y_{1i}+y_{1i}}{2}\right)$

Here, $y_{CM.f}$is the final and$y_{CM.i}$is the initial center of mass in the $y$direction and$y_{1f}\mathrm{and}{\mathrm{y}}_{1\mathrm{i}}$are the final and the initial center of mass in the first case and $y_{2i}$and $y_{2f}$are the initial and the final center of mass in the second case.

The above equation (i) can also be written as:

$∆K_{translational}=F_{net}\left(y_{CM.f}-y_{CM.i}\right)$

Substitute the values of equation (ii) and equation (iii) in the above equation.

$∆K_{translationa{l}_{}}=\left(F-F_{grav}\right)\left(\frac{y_{1f}+y_{2f}}{2}-\frac{y_{1i}+y_{2i}}{2}\right)$

Substitute all the values in the above expression.

$$\begin{gathered} ∆K_{translationa{l}_{}}=\left(167\mathrm{N}-2×5\mathrm{kg}×9.8\mathrm{m}/\mathrm{s}\right)\left(\frac{0.5\mathrm{m}+1.2\mathrm{m}}{2}-\frac{0.3\mathrm{m}+0.7\mathrm{m}}{2}\right) \\ =\left(167\mathrm{N}-98\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\frac{1\mathrm{N}}{1\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2}\right)\left(\frac{1.7\mathrm{m}}{2}-\frac{1\mathrm{m}}{2}\right) \\ =69\mathrm{N}×0.35\mathrm{m} \\ =24.15\mathrm{N}.\mathrm{m}×\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}} \\ =24.15\mathrm{J} \end{gathered}$$

The equation of the energy of the real system can be expressed as:

$∆E_1=W+Q$

Here,$W$is the work done by the system and$Q$is the heat added to the system.

The equation of the increase in the total energy can be expressed as:

$∆K_{total}=\underset{}{∑{\stackrel{→}{F}}_1∆r_1+∆{\stackrel{→}{F}}_2∆r_2+......}$

Here, $∆K_{total}$is the total kinetic energy and$\underset{}{∑{\stackrel{→}{F}}_1}$is the summation of all the acting forces.

Using the value of the other kinetic energies, the values of the above equation can be expressed as:

$$\begin{gathered} ∆K_{translation}+∆K_{vibration}+∆{\mathrm{U}}_{\mathrm{spring}}=\mathrm{F}\left({\mathrm{y}}_{2\mathrm{f}}-{\mathrm{y}}_{2\mathrm{i}}\right)-2\mathrm{Mg}\left({\mathrm{y}}_{\mathrm{CM},\mathrm{f}}-{\mathrm{y}}_{\mathrm{CM},\mathrm{i}}\right) \\ ∆{\mathrm{K}}_{\mathrm{vibration}}+∆{\mathrm{U}}_{\mathrm{spring}}=\mathrm{F}\left({\mathrm{y}}_{2\mathrm{f}}-{\mathrm{y}}_{2\mathrm{i}}\right)-2\mathrm{Mg}\left(\frac{{\mathrm{y}}_{1\mathrm{f}}-{\mathrm{y}}_{2\mathrm{i}}}{2}-\frac{{\mathrm{y}}_{1\mathrm{i}}-{\mathrm{y}}_{2\mathrm{i}}}{2}\right)-∆{\mathrm{K}}_{\mathrm{translation}} \end{gathered}$$

Here, $∆K_{translation}$is the translational kinetic energy, $∆K_{vibration}$is the vibrational energy and $∆{\mathrm{U}}_{\mathrm{spring}}$is the potential energy

Substituting all the values in the above equation.

$$\begin{gathered} ∆K_{translation}+∆{\mathrm{U}}_{\mathrm{spring}}=167\mathrm{N}\left(1.2\mathrm{m}-0.7\mathrm{m}\right)-2×5\mathrm{kg}×9.8\mathrm{m}/{\mathrm{s}}^2\left(\frac{0.5\mathrm{m}+0.7\mathrm{m}}{2}-\frac{0.3\mathrm{m}+0.7\mathrm{m}}{2}\right)-∆{\mathrm{K}}_{\mathrm{translation}} \\ =83.5\mathrm{N}.\mathrm{m}×\frac{1\mathrm{J}}{1\mathrm{N}.\mathrm{m}}-98\mathrm{kg}.\mathrm{m}/{\mathrm{s}}^2\left(0.35\mathrm{m}\right)-24.15\mathrm{J} \\ =83.5\mathrm{J}-34.3\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2×\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}-24.15\mathrm{J} \\ =25.5\mathrm{J} \end{gathered}$$

Thus, the increase of the vibrational kinetic energy plus the potential energy of the spring is $25.5\mathrm{J}$.