Chapter 6, Problem 57QAP

To determine

(a)

Speed of the object, when it is $8.00$ cm from equilibrium

Answer to Problem 57QAP

Speed of the object, when it is $8.00$ cm from equilibrium= $0.42{\text{ ms}}^{\text{-1}}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Object mass= }5.00\text{ kg}\\ \text{spring constant = }250{\text{ Nm}}^{\text{-1}}\\ \text{spring is compressed 10 cm}\end{array}$

Formula used:

  $\begin{array}{l}E=\frac{1}{2}k{x}^2\\ E=\text{energy}\\ k=\text{spring constant}\\ x=\text{distance}\end{array}$

  $\begin{array}{l}E=\frac{1}{2}m{v}^2\\ E=\text{kinetic energy}\\ m=\text{mass}\\ v=\text{velocity}\end{array}$

Calculation:

  $\begin{array}{l}\text{Potential energy of the spring when it is compressed 10}\text{.0 cm,}\\ E=\frac{1}{2}k{x}^2\\ E=\frac{1}{2}\times 250{\text{ Nm}}^{\text{-1}}\times {\text{(}}^{0.1}\\ E=1.25\text{ J}\\ \text{When released, potential energy is converted to kinetic energy}.\end{array}$

  $\begin{array}{l}\text{As the total energy is constant, when it is at 8 cm,}\\ E_k=E_{p10}-E_{p8}\\ E_k=\text{kinetic energy}\\ E_{p8}=\text{potential energy at 8 cm}\\ E_{p10}=\text{potential energy at 10 cm}\\ E_k=1.25\text{ J}-\frac{1}{2}*250{\text{ Nm}}^{\text{-1}}\text{*(}0.08{\text{ m)}}^2\\ E_k=0.45\text{ J}\end{array}$

  $\begin{array}{l}E=\frac{1}{2}m{v}^2\\ 0.45\text{ J=}\frac{1}{2}\times 5\text{ kg}\times v^2\\ 0.45\text{ J=}\frac{1}{2}\times 5\text{ kg}\times v^2\\ v=0.42{\text{ ms}}^{\text{-1}}\end{array}$

Conclusion:

Speed of the object, when it is $8.00$ cm from equilibrium= $0.42{\text{ ms}}^{\text{-1}}$

To determine

(b)

Speed of the object, when it is $5.00$ cm from equilibrium

Answer to Problem 57QAP

Speed of the object, when it is $5.00$ cm from equilibrium= $0.61{\text{ ms}}^{\text{-1}}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Object mass= }5.00\text{ kg}\\ \text{spring constant = }250{\text{ Nm}}^{\text{-1}}\\ \text{spring is compressed 10 cm}\end{array}$

Formula used:

  $\begin{array}{l}E=\frac{1}{2}k{x}^2\\ E=\text{energy}\\ k=\text{spring constant}\\ x=\text{distance}\end{array}$

  $\begin{array}{l}E=\frac{1}{2}m{v}^2\\ E=\text{kinetic energy}\\ m=\text{mass}\\ v=\text{velocity}\end{array}$

Calculation:

  $\begin{array}{l}\text{Potential energy of the spring when it is compressed 10}\text{.0 cm,}\\ E=\frac{1}{2}k{x}^2\\ E=\frac{1}{2}\times 250{\text{ Nm}}^{\text{-1}}\times {\text{(}}^{0.1}\\ E=1.25\text{ J}\\ \text{When released, potential energy is converted to kinetic energy}.\\ \text{As the total energy is constant, when it is at 5 cm,}\\ E_k=E_{p10}-E_{p5}\\ E_k=\text{kinetic energy}\\ E_{p5}=\text{potential energy at 5 cm}\\ E_{p10}=\text{potential energy at 10 cm}\\ E_k=1.25\text{ J}-\frac{1}{2}\times 250{\text{ Nm}}^{\text{-1}}\times {\text{(}}^{0.05}\\ E_k=0.94\text{ J}\end{array}$

  $\begin{array}{l}E=\frac{1}{2}m{v}^2\\ E_k=0.94\text{ J=}\frac{1}{2}\times 5\text{ kg}\times v^2\\ 0.94\text{ J=}\frac{1}{2}\times 5\text{ kg}\times v^2\\ v=0.61{\text{ ms}}^{\text{-1}}\end{array}$

Conclusion:

Speed of the object, when it is $5.00$ cm from equilibrium= $0.61{\text{ ms}}^{\text{-1}}$

To determine

(c)

Speed of the object, when it is at equilibrium

Answer to Problem 57QAP

Speed of the object, when it is at equilibrium= $0.71{\text{ ms}}^{\text{-1}}$

Explanation of Solution

Given info:

  $\begin{array}{l}\text{Object mass= }5.00\text{ kg}\\ \text{spring constant = }250{\text{ Nm}}^{\text{-1}}\\ \text{spring is compressed 10 cm}\end{array}$

Formula used:

  $\begin{array}{l}E=\frac{1}{2}k{x}^2\\ E=\text{energy}\\ k=\text{spring constant}\\ x=\text{distance}\end{array}$

  $\begin{array}{l}E=\frac{1}{2}m{v}^2\\ E=\text{kinetic energy}\\ m=\text{mass}\\ v=\text{velocity}\end{array}$

Calculation:

  $\begin{array}{l}\text{Potential energy of the spring when it is compressed 10}\text{.0 cm,}\\ E=\frac{1}{2}k{x}^2\\ E=\frac{1}{2}\times 250{\text{ Nm}}^{\text{-1}}\times {\text{(}}^{0.1}\\ E=1.25\text{ J}\\ \text{When released, at equilibrium all the potential energy is converted to kinetic energy}.\end{array}$

  $\begin{array}{l}E=\frac{1}{2}m{v}^2\\ E_k=1.25\text{ J=}\frac{1}{2}\times 5\text{ kg}\times v^2\\ 0.94\text{ J=}\frac{1}{2}\times 5\text{ kg}\times v^2\\ v=0.71{\text{ ms}}^{\text{-1}}\end{array}$

Conclusion:

Speed of the object, when it is at equilibrium= $0.71{\text{ ms}}^{\text{-1}}$