91Ó°ÊÓ

The mass of the block is, m = 20 kg

The spring constant of the spring is,$\mathrm{k}=4.0\mathrm{kN}/\mathrm{m}\left(\frac{{10}^3\mathrm{N}/\mathrm{m}}{1\mathrm{kN}/\mathrm{m}}\right)=4.0×{10}^3\mathrm{N}/\mathrm{m}$

The stretched length of the block is,$×=10\mathrm{cm}\left(\frac{1\mathrm{m}}{100\mathrm{cm}}\right)=0.10\mathrm{m}$

The magnitude of the force is,$\mathrm{F}=80\mathrm{N}$

We can use the law of conservation of energy to solve this. It states that the sum of potential and kinetic energy is equal to mechanical energy and this always remains constant.

Formulae:

The kinetic energy of a body while stretching, $∆\mathrm{K}={∫}_{{\mathrm{x}}_1}^{{\mathrm{x}}_2}=-\mathrm{kxdx}$ (1)

Using equation (1), we can get the change in the kinetic energy of the block as given:

$$\begin{gathered} ∆\mathrm{K}={∫}_{0.10}^{0.08}-\mathrm{kxdx} \\ =-0.5×\mathrm{k}×{\left[{\mathrm{x}}^2\right]}_{0.01}^{0.08} \\ =-0.5×4.0×{10}^3\mathrm{N}/\mathrm{m}×\left({\left(0.08\mathrm{m}\right)}^2-{\left(0.1\mathrm{m}\right)}^2\right) \\ =-7.2\mathrm{J} \end{gathered}$$

Now, applying the law of conservation of energy, we can get the kinetic energy of the block as:

$$\begin{gathered} \mathrm{K}+∆\mathrm{K}=-{\mathrm{f}}_{\mathrm{r}}×\mathrm{x} \\ \mathrm{K}-7.2\mathrm{J}=-80\mathrm{N}×0.02\mathrm{m} \\ \mathrm{K}=5.6\mathrm{J} \end{gathered}$$

Hence, the value of the energy is 5.6 J.

Using equation (1), the kinetic energy of the block when it slides back to that point is given as:

Hence, the kinetic energy of the block is 12 J.

The net displacement is given by:

$$\begin{gathered} \mathrm{x}=0.10\mathrm{m}-\mathrm{d} \\ 0.02\mathrm{m}=0.10\mathrm{m}-\mathrm{d} \\ \mathrm{d}=0.08\mathrm{m} \end{gathered}$$

The change in kinetic energy is given using equation (i) as follows:

$$\begin{gathered} ∆\mathrm{K}={∫}_{0.10}^{0.02}-\mathrm{kxdx} \\ =-0.5×\mathrm{k}×{\left[{\mathrm{x}}^2\right]}_{0.01}^{0.02} \\ =-0.5×4×{10}^3\mathrm{N}/\mathrm{m}×\left[{\left(0.02\mathrm{m}\right)}^2-{\left(0.01\mathrm{m}\right)}^2\right] \\ =-19.2\mathrm{J} \end{gathered}$$

Now applying the law of conservation of energy, the maximum energy obtained by the block is given as:

$$\begin{gathered} {\mathrm{K}}_{\mathrm{ma}}+∆\mathrm{K}=-{\mathrm{f}}_{\mathrm{r}}×\mathrm{d} \\ {\mathrm{K}}_{\mathrm{ma}}-19.2\mathrm{J}=-80\mathrm{N}×0.08\mathrm{m} \\ {\mathrm{K}}_{\mathrm{ma}}=12.8\mathrm{J} \end{gathered}$$

Hence, the value of the maximum energy is 12.8 J.