91Ó°ÊÓ

The given data can be listed below as,

• The stiffness of the spring is,$k=200N/m$ .
• The mass of the spring is,role="math" localid="1668488385102" $m=0.4kg$.
• The initial compression of the spring is,$x_c=10 \text{cm}$.
• The speed of the mass is,$v_i=3m/s$.

A body's stiffness is a measurement of how resistant an elastic body is to deforming. In general, the force necessary to generate unit deformation for a spring is determined by the spring stiffness.

The expression for the conservation of energy is expressed as,

$\frac{1}{2}k{x}_c^2+\frac{1}{2}m{v}_i^2=\frac{1}{2}k{x}_m^2+\frac{1}{2}m{v}_f^2$

Here, $k$is the spring constant,$x_c$ is the initial compression,$m$ is the mass,$x_m$ is the maximum stretch,$v_i$ is the initial speed and$v_f$ is the final speed.

Substitute all the value in the above equation.

$$\begin{gathered} \frac{1}{2}\left(200N/m\right)×{\left(0.1m\right)}^2+\frac{1}{2}\left(0.4kg\right)×{\left(3m/s\right)}^2=\frac{1}{2}\left(200N/m\right)x_m^2+\frac{1}{2}\left(0.4kg\right)×0 \\ {x}_m=0.167m \end{gathered}$$

Hence the maximum stretch is $0.167 \text{m}$.

The conservation of energy between the initial and final states is expressed as,

$\frac{1}{2}k{x}_c^2+\frac{1}{2}m{v}_i^2=\frac{1}{2}m{v}_m^2$

Substitute all the value in the above equation.

$$\begin{gathered} \frac{1}{2}\left(200N/m\right)×{\left(0.1m\right)}^2+\frac{1}{2}\left(0.4kg\right)×{\left(3m/s\right)}^2=\frac{1}{2}\left(0.4kg\right)×v_m^2 \\ {v}_m=3.74m/s \end{gathered}$$

Hencethe maximum speed is,$3.74m/s$.

Consider each cycle has energy dissipation is, $0.01 \text{J}$

The energy must be supply steady state oscillation is $0.01 \text{J}$per cycle.

The time period per cycle of per cycle is,

$$\begin{gathered} T=\frac{2\mathrm{Ï€}}{\mathrm{Ó\neg }} \\ T=2\mathrm{Ï€}\sqrt{\frac{m}{k}} \end{gathered}$$

Substitute all the value in the above equation.

$$\begin{gathered} T=2\mathrm{Ï€}\sqrt{\frac{0.4\mathrm{kg}}{200\mathrm{N}/\mathrm{m}}} \\ \mathrm{T}=0.28\mathrm{s} \end{gathered}$$

Average input energy per second is,

$$\begin{gathered} \frac{\mathrm{Δ}E}{T}=\frac{0.01 \text{J}}{0.28 \text{s}} \\ =0.036J/s \\ =0.036watts \end{gathered}$$

Hence the average input energy per second is, $0.036watts$.