Question

A spring with spring constant k=200 N/mis suspended vertically with its upper end fixed to the ceiling and its lower end at position y=0. A block of weight 20 Nis attached to the lower end, held still for a moment, and then released. What are (a) the kinetic energy, (b) the change (from the initial value) in the gravitational potential energy$\mathbf{∆}{\mathit{U}}_{\mathbf{g}}$, and (c) the change in the elastic potential energy$\mathbf{∆}{\mathit{U}}_{\mathbf{e}}$of the spring–block system when the block is at y=5 cm? What are (d) K, (e)$\mathbf{∆}{\mathit{U}}_{\mathbf{g}}$, and (f)$\mathbf{∆}{\mathit{U}}_{\mathbf{e}}$when y=10cm, (g) K, (h)$\mathbf{∆}{\mathit{U}}_{\mathbf{g}}$, and (i)$\mathbf{∆}{\mathit{U}}_{\mathbf{g}}$when y=15cm, and (j) K, (k)$\mathbf{∆}{\mathit{U}}_{\mathbf{g}}$, and (l)$\mathbf{∆}{\mathit{U}}_{\mathbf{e}}$when y=20cm?

Short answer

a) Kinetic energy at $∆y=-5cm$ is0.75J.

b) Change in gravitational potential energy at $∆y=-5cm$ is-1.0J.

c) Change in elastic potential energy at $\Delta y=-5cm$ is0.25J.

d) Kinetic energy at $\Delta y=-10cm$ is1.0J.

e) Change in gravitational potential energy at $\Delta y=-10cm$is-2.0J.

f) Change in elastic potential energy at $\Delta y=-10cm$is1.0J.

g) Kinetic energy at$\Delta y=-15cm$ is0.75J.

h) Change in gravitational potential energy at$\Delta y=-15cm$ is-3.0J.

i) Change in elastic potential energy at$\Delta y=-15cm$ is2.25J.

j) Kinetic energy at$\Delta y=-20cm$ is0 J.

k) Change in gravitational potential energy at$\Delta y=-20cm$ is-4.0 J.

l) Change in elastic potential energy at$\Delta y=-20cm$ is4.0J.

Step-by-step solution

The given data

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

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

Understanding the concept of energy

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,$U_g=mg\Delta y$ (ii)

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

a) Calculation of kinetic energy for ∆y=5cm

For$\Delta y=-0.05m$

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:

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

Hence, the value of the energy is 0.75 J.

b) Calculation of gravitational potential energy for ∆y=5cm

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

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

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

c) Calculation of elastic potential energy for ∆y=5cm

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

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

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

d) Calculation of kinetic energy for ∆y=10cm

For$\Delta y=-10cm$

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:

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

Hence, the value of the energy is 1.0J.

e) Calculation of gravitational potential energy for∆y=10cm

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

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

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

f) Calculation of elastic potential energy for∆y=10cm

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

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

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

g) Calculation of kinetic energy for∆y=15cm

For$\Delta y=-0.15m$

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

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

Hence, the value of the energy is 0.75J.

h) Calculation of gravitational potential energy for ∆y=15cm

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

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

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

i) Calculation of elastic potential energy for Δy=15cm

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

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

Hence, the value of the energy is 2.25J.

j) Calculation of kinetic energy forΔy=20cm

For$\Delta y=-0.2m$

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:

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

Hence, the value of the energy is 0.0J.

k) Calculation of gravitational potential energy forΔy=20cm

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

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

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

l) Calculation of elastic potential energy forΔy=20cm

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

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

Hence, the value of the energy is 4.0 J.