91Ó°ÊÓ

i) Spring stiffness is$k=200\mathrm{N}/\mathrm{m}$

ii) Weight of block is$W=20N$

We have to use conservation of energy in which the summation of kinetic energy, change in potential energy, and change in elastic energy is zero.

Formulae:

The kinetic energy of a body,$K=0.5m{v}^2$ (i)

The gravitational potential energy of a body,$U_g=mg\mathrm{Δ}y$ (ii)

The elastic potential energy of a spring, $U_e=0.5k\mathrm{Δ}{y}^2$ (iii)

For$\mathrm{Δ}y=-0.05m$

From the lawof conservation of energy, total energy at bottom is conserved. Thus, the kinetic energy is given using equations (ii) and (iii) as follows:

$$\begin{gathered} E_{bottom}=0 \\ K+∆U_g+∆U_e=0 \\ K+W∆y+0.5×k×∆y^2=0 \\ K+20×\left(-0.05\right)+0.5×200×{\left(-0.05\right)}^2=0 \\ K=0.75J \end{gathered}$$

Hence, the value of the energy is 0.75 J.

The change in gravitational energy is given using equation (ii) as:

$$\begin{gathered} ∆U_g=20×\left(-0.05\right) \\ =1.0J \end{gathered}$$

Hence, the value of the energy is-1.0 J.

The change in elastic potential energy is given using equation (iii) as follows:

$$\begin{gathered} ∆U_e=0.5×200x{\left(-0.05\right)}^2 \\ =0.25J \end{gathered}$$

Hence, the value of the energy is0.25J.

For$\mathrm{Δ}y=-10cm$

From law of conservation of energy, the total energy at bottom is conserved. Thus, the kinetic energy using equations (ii) and (iii) is given as:

$$\begin{gathered} E_{bottom}=0 \\ K+∆U_g+∆U_e=0 \\ K+W∆y+0.5×k×∆y^2=0 \\ K+20×\left(-0.1\right)+0.5×200×{\left(-0.1\right)}^2=0 \\ K=1.0J \end{gathered}$$

Hence, the value of the energy is 1.0J.

The change in gravitational energy is given using equation (ii) as:

$$\begin{gathered} ∆U_g=20×\left(-0.1\right) \\ =-2.0J \end{gathered}$$

Hence, the value of the energy is-2.0J.

The change in elastic potential energy is given using equation (iii) as follows:

$$\begin{gathered} ∆U_e=0.5×200x{\left(-0.1\right)}^2 \\ =1.0J \end{gathered}$$

Hence, the value of the energy is1.0J.

For$\mathrm{Δ}y=-0.15m$

From the lawof conservation of energy, the total energyat the bottomis conserved. Thus, the kinetic energy using equations (ii) and (iii) is given as:

$$\begin{gathered} E_{bottom}=0 \\ K+∆U_g+∆U_e=0 \\ K+W∆y+0.5×k×∆y^2=0 \\ K+20×\left(-0.15\right)+0.5×200×{\left(-0.15\right)}^2=0 \\ K=0.75J \end{gathered}$$

Hence, the value of the energy is 0.75J.

The change in gravitational energy is given using equation (ii) as follows:

$$\begin{gathered} ∆U_g=20×\left(-0.15\right) \\ =-3.0J \end{gathered}$$

Hence, the value of the energy is -3.0J.

The change in elastic potential energy is given using equation (iii) as follows:

$$\begin{gathered} ∆U_e=0.5×200x{\left(-0.15\right)}^2 \\ =2.25J \end{gathered}$$

Hence, the value of the energy is 2.25J.

For$\mathrm{Δ}y=-0.2m$

Kinetic energy is calculated as follows:

From the lawof conservation of energy, the total energyat the bottomis conserved. Thus, the kinetic energy is given using equations (ii) and (iii) as follows:

$$\begin{gathered} E_{bottom}=0 \\ K+∆U_g+∆U_e=0 \\ K+W∆y+0.5×k×∆y^2=0 \\ K+20×\left(-0.2\right)+0.5×200×{\left(-0.2\right)}^2=0 \\ K=0.0J \end{gathered}$$

Hence, the value of the energy is 0.0J.

The change in gravitational energy is given using equation (ii) as follows:

$$\begin{gathered} ∆U_g=20×\left(-0.20\right) \\ =-4.0J \end{gathered}$$

Hence, the value of the energy is -4.0 J.

The change in elastic potential energy is given using equation (iii) as follows:

$$\begin{gathered} ∆U_e=0.5×200x{\left(-0.2\right)}^2 \\ =4.0J \end{gathered}$$

Hence, the value of the energy is 4.0 J.