Chapter 12, Problem 78QAP

To determine

(a)

Derive the formula for speed using conversion of energy. and using the formula calculate the speed of the object.

Answer to Problem 78QAP

  $v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

Explanation of Solution

By conservation of energy,
Spring potential energy when maximum compression=spring PE+KE

  $\begin{array}{l}\frac{1}{2}k{A}^2=\frac{1}{2}k{x}^2+\frac{1}{2}m{v}^2\\ m{v}^2=k{A}^2-k{x}^2\\ v^2=\frac{k\left(A^2-x^2\right)}{m}\\ v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}\end{array}$

To determine

(b)

The speed of the object.

Answer to Problem 78QAP

At $x=0$ cm; $v=1.843m/s$

At $x=2$ cm; $v=1.806m/s$

At $x=5$ cm; $v=1.820m/s$

At $x=8$ cm; $v=1.106m/s$

At $x=10$ cm; $v=1.834m/s$

Explanation of Solution

Given:

M= $250g=0.25kg$$$

K= $85N/m$

A= $10cm$

X= $0cm,2cm,5cm,8cm,10cm$

Calculation:

At $x=0$cm
  $v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

  $\begin{array}{l}v=\sqrt{\frac{85\left({\left(0.1\right)}^2-{\left(0\right)}^2\right)}{0.25}}\\ v=1.843m/s\end{array}$

At $x=2$cm
  $v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

  $\begin{array}{l}v=\sqrt{\frac{85\left({\left(0.1\right)}^2-{\left(0.02\right)}^2\right)}{0.25}}\\ v=1.806m/s\end{array}$ ).

At $x=5$cm
  $v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

  $\begin{array}{l}v=\sqrt{\frac{85\left({\left(0.1\right)}^2-{\left(0.05\right)}^2\right)}{0.25}}\\ v=1.820m/s\end{array}$ ).

At $x=8$cm
  $v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

  $\begin{array}{l}v=\sqrt{\frac{85\left({\left(0.1\right)}^2-{\left(0.08\right)}^2\right)}{0.25}}\\ v=1.106m/s\end{array}$ ).

At $x=10$cm
  $v=\sqrt{\frac{k\left(A^2-x^2\right)}{m}}$

  $\begin{array}{l}v=\sqrt{\frac{85\left({\left(0.1\right)}^2-{\left(0.01\right)}^2\right)}{0.25}}\\ v=1.834m/s\end{array}$