Chapter 6, Problem 32P

To determine

ToDetermine:Height that the person Jswings upward.

Answer to Problem 32P

Solution: $1.275\text{ m}$

Explanation of Solution

Given:$5.0\text{ m/s}$

Formula Used:

The forces acting on Person J are the force due to gravity and tension due to the vine. Work done by the tension force is zero because of perpendicular to the direction of her motion.

According to the law of conservation of energy:

  $\begin{array}{l}{E}_{initial}=E_{final}\\ mg{h}_i+\frac{1}{2}m{v}_1{}^2=mg{h}_f+\frac{1}{2}m{v}_f{}^2\end{array}$

Where m is the mass of Person J, $h_i$ is the initial height of Person J, $v_i$ is the initial velocity of Person J, $v_f$ is the final velocity of the Person J, $h_f$ is the final height of the Person J, g is acceleration due to gravity.

Calculations:

She has kinetic energy, but potential energy is zero and she is having swung up to some height h, and momentarily at rest. Here she has potential energy, and kinetic energy is zero.

Substituting velocity value in above equation and solve for final height

  $\begin{array}{l}mg{h}_i+\frac{1}{2}m{v}_1{}^2=mg{h}_f+\frac{1}{2}m{v}_f{}^2\\ 0+\frac{1}{2}m{v}_1{}^2=mg{h}_f+0\\ \frac{1}{2}m{v}_1{}^2=mg{h}_f\\ \frac{1}{2\times g}{v}_1{}^2=h_f\\ h_f=\frac{1}{2\times 9.8}{5}^2\\ h_f=1.275\text{ m}\end{array}$

Person J can swing upward of height $1.275\text{ m}\text{.}$

Conclusion:Person J can swing upward of height $1.275\text{ m}$ to look for Tarzan,Person J running along the level ground with a velocity of $5.0\text{ m/s}.$