Chapter 6, Problem 94P

(a)

To determine

The spring constant of the spring.

Answer to Problem 94P

The spring constant of the spring is $\underset{\_}{24.83\text{N}/\text{m}}$.

Explanation of Solution

According to law of conservation of energy.

  $\begin{array}{c}{U}_{plastic}=U_{elastic}\\ \frac{1}{2}k{x}^2=mg\Delta h\end{array}$        (I)

Here, $k$ is the spring constant, $x$ is the displacement, $m$ is the mass, $g$ is the acceleration due to gravity, and $\Delta h$ is the height.

Rearrange equation (I) to obtain an expression for $k$.

  $k=\frac{2mg\Delta h}{x^2}$        (II)

Conclusion:

Substitute, $780\text{N}$ for $mg$, $\Delta h-30\text{m}$ for $x$, and $182\text{m}-67\text{m}$ for $\Delta h$, in equation (II) to find $k$.

  $\begin{array}{c}k=\frac{2\left(780\text{N}\right)\left(182\text{m}-67\text{m}\right)}{{\left(182\text{m}-67\text{m}-30\text{m}\right)}^2}\\ =24.83\text{N}/\text{m}\end{array}$

Therefore, the spring constant of the spring is $\underset{\_}{24.83\text{N}/\text{m}}$.

(b)

To determine

The speed of the jumper when he reaches height of $92\text{m}$ above the bottom of the gorge.

Answer to Problem 94P

The speed of the jumper when he reaches height of $92\text{m}$ above the bottom of the gorge is $\underset{\_}{23.76\text{m}/\text{s}}$.

Explanation of Solution

Write the expression for potential energy.

  $U_{potential}=mg\Delta h$        (III)

Here, $m$ is the mass, $g$ is the acceleration due to gravity, $\Delta h$ is the height.

Write the expression for energy of the spring.

  $U_{spring}=\frac{1}{2}k{x}^2$        (IV)

Write the expression for change in energy.

  $\Delta k=U_{potential}-U_{elastic}$        (V)

Write the expression for kinetic energy.

  $\Delta k=\frac{1}{2}m{\upsilon }^2$        (VI)

Here, $\upsilon$ is the speed.

Rearrange equation (VI) to obtain an expression for $\upsilon$.

  $\upsilon =\sqrt{\frac{2\Delta k}{m}}$        (VII)

Conclusion:

Substitute, $780\text{N}$ for $mg$, and $182\text{m}-\text{92}\text{m}$ for $\Delta h$ in equation (III).

  $\begin{array}{c}{U}_{potential}=\left(780\text{N}\right)\left(182\text{m}-\text{93}\text{m}\right)\\ =70200\text{J}\end{array}$

Substitute, $24.83\text{N}/\text{m}$ for $k$, and $92\text{m}-\text{30}\text{m}$ for $x$ in equation (IV).

  $\begin{array}{c}{U}_{spring}=\frac{1}{2}\left(24.83\text{N}/\text{m}\right){\left(92\text{m}-\text{30}\text{m}\right)}^2\\ =47723\text{J}\end{array}$

Substitute, $70200\text{J}$ for $U_{potential}$, $47723\text{J}$ $U_{spring}$ in equation (V).

  $\begin{array}{c}\Delta k=70200\text{J}-47723\text{J}\\ =22477\text{J}\end{array}$

Substitute, $79.6\text{kg}$ for $m$, and $22477\text{J}$ for $\Delta k$ in equation (VII).

  $\begin{array}{c}\upsilon =\sqrt{\frac{2\left(22477\text{J}\right)}{79.6\text{kg}}}\\ =23.76\text{m}/\text{s}\end{array}$

Therefore, the speed of the jumper when he reaches height of $92\text{m}$ above the bottom of the gorge is $\underset{\_}{23.76\text{m}/\text{s}}$.