Chapter 6, Problem 44P

(a)

To determine

The maximum speed.

Answer to Problem 44P

Solution:

  $\sqrt{v_0{}^2+\frac{k}{m}{\left(x_0\right)}^2}$

Explanation of Solution

Given:

Spring stiffness constant $=k$

Mass of block $=m$

Initial velocity $\left(v_1\right)=v_0$

Initial elongation of spring $=x_0$

Formula Used:

In the absence of friction force total mechanical energy is conserved.

  $\text{K}{\text{.E}}_{\text{1}}\text{+P}{\text{.E}}_{\text{1}}\text{=K}{\text{.E}}_{\text{2}}\text{+P}{\text{.E}}_{\text{2}}$

Initial potential energy: $P{E}_1$

Final potential energy: $P{E}_2$

Initial kinetic energy: $K{E}_1$

Final kinetic energy: $K{E}_2$

Elastic potential energy of spring $=\frac{1}{2}k{\left(x\right)}^2$

Where, k : spring stiffness constant

x : elongation or compression from natural length.

Calculation:

Initial K.E:

  $\begin{array}{l}\left(K.E_1\right)=\frac{1}{2}m{v}_1{}^2\\ =\frac{1}{2}m{v}_0{}^2\end{array}$

Initial elastic P.E of spring $\left(\text{P}{\text{.E}}_{\text{1}}\right)$

  $=\frac{1}{2}k{x}_0{}^2$

Maximum speed of block is only when, there is neither compression nor elongation in spring

That is spring potential energy should be zero.

Final elastic P.E of spring

  $\begin{array}{c}\left(\text{P}{\text{.E}}_{\text{2}}\right)\text{=}\frac{\text{1}}{\text{2}}k{\left(\text{0}\right)}^{\text{2}}\\ \text{=0J}\end{array}$

Final kinetic energy of block $\text{=}\frac{1}{2}m{v}_{\max }{}^2$

Apply conservation of mechanical energy

  $K.E_1+P.E_1=K.E_2+P.E_2$

  $\frac{1}{2}m{v}_0{}^2+\frac{1}{2}k{x}_0{}^2=\frac{1}{2}m{v}_{\max }{}^2+0$

  $\begin{array}{l}{v}_{\max }=\sqrt{\frac{m{v}_0{}^2+k{x}_0{}^2}{m}}\\ v_{\max }=\sqrt{v_0{}^2+\frac{k}{m}{\left(x_0\right)}^2}\end{array}$

Conclusion: Thus, the maximum speed would be: $\sqrt{v_0{}^2+\frac{k}{m}{\left(x_0\right)}^2}$

(b)

To determine

The maximum stretch from equilibrium in the term of given quantities.

Answer to Problem 44P

Solution:

Maximum elongation of spring $\left(x_{\max }\right)$

  $=\sqrt{x_0{}^2+\frac{m}{k}{\left(v_0\right)}^2}$

Explanation of Solution

Given:

Spring stiffness constant $=k$

Mass of block $=m$

Initial velocity $\left(v_1\right)=v_0$

Initial elongation of spring $=x_0$

Formula Used:

In the absence of friction force total mechanical energy is conserved.

  $\text{K}{\text{.E}}_{\text{1}}\text{+P}{\text{.E}}_{\text{1}}\text{=K}{\text{.E}}_{\text{2}}\text{+P}{\text{.E}}_{\text{2}}$

Initial potential energy: $P{E}_1$

Final potential energy: $P{E}_2$

Initial kinetic energy: $K{E}_1$

Final kinetic energy: $K{E}_2$

Elastic potential energy of spring $=\frac{1}{2}k{\left(x\right)}^2$

Where, k : spring stiffness constant

x : elongation or compression from natural length.

Calculation:

Initial K.E $\left(\text{K}{\text{.E}}_{\text{1}}\right)\text{=}\frac{1}{2}m{v}_1{}^2$

  $\text{=}\frac{1}{2}m{v}_0{}^2$

Initial elastic P.E of spring $\left(\text{P}{\text{.E}}_{\text{1}}\right)$

  $\text{=}\frac{1}{2}k{x}_0$

For maximum elongation in spring block should be at rest

Final elastic P.E of spring $\left(\text{P}{\text{.E}}_{\text{2}}\right)$

  $\text{=}\frac{1}{2}k{x}_{\max }{}^2$

Final K.E of block $=\frac{1}{2}m{v}_2{}^2$

  $\begin{array}{l}=\frac{1}{2}m{\left(\text{0}\right)}^{\text{2}}\\ =0\text{J}\end{array}$

Apply conservation of energy

  $\text{K}{\text{.E}}_{\text{1}}\text{+P}{\text{.E}}_{\text{1}}\text{=K}{\text{.E}}_{\text{2}}\text{+P}{\text{.E}}_{\text{2}}$

  $\begin{array}{l}\frac{1}{2}m{v}_0{}^2+\frac{1}{2}k{x}_0{}^2=\frac{1}{2}k{x}_{\max }{}^2+0\\ x_{\max }=\sqrt{x_0{}^2+\frac{m}{k}{v}_0{}^2}\end{array}$

Conclusion: Thus, the maximum elongation of spring $\left(x_{\max }\right)$

  $=\sqrt{\left(x_0{}^2+\frac{m}{k}{v}_0{}^2\right)}$ .