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Chapter 7: Problem 12

Tarzan and Jane. Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 \(\mathrm{m}\) that makes an angle of \(45^{\circ}\) with the vertical, steps off his tree limb, and swings down and then up to Jane's open arms. When he arrives, his vine makes an angle of \(30^{\circ}\) with the vertical. Determine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

Short Answer

Expert verified
Calculate the final expression to get Tarzan's final speed. If it's not exceedingly high, he likely reaches Jane safely.

Step by step solution

01

Determine the initial and final heights

Tarzan swings from one tree to another using a vine. First, determine the heights relative to the lowest point of Tarzan's swing. The vine is 20 meters long. Using trigonometry, the height at the start (initial) when the vine makes a 45掳 angle is: \[ h_i = 20 \cos(45掳) = 20 \times \frac{\sqrt{2}}{2} = 10 \sqrt{2}\] For the final position, when the vine makes a 30掳 angle, the height is: \[ h_f = 20 \cos(30掳) = 20 \times \frac{\sqrt{3}}{2} = 10\sqrt{3}\]
02

Calculate change in potential energy

Next, find the change in gravitational potential energy (PE), which will be converted into kinetic energy (KE). The change in height is:\[ \Delta h = h_i - h_f = 10 \sqrt{2} - 10 \sqrt{3}\]Potential energy difference is: \[ \Delta PE = mg\Delta h = mg(10 \sqrt{2} - 10 \sqrt{3})\] where \(m\) is Tarzan's mass and \(g\) is acceleration due to gravity (9.8 m/s虏).
03

Convert potential energy to kinetic energy

Assuming no energy losses, change in PE is converted to KE as Tarzan swings:\[ \Delta PE = \Delta KE = \frac{1}{2}mv^2\]Rearranging gives:\[ v^2 = 2g (10 \sqrt{2} - 10 \sqrt{3})\] From this, find Tarzan's speed \(v\):\[ v = \sqrt{2g (10 \sqrt{2} - 10 \sqrt{3})}\]
04

Calculate Tarzan's speed before reaching Jane

Substitute the known value of \(g = 9.8\,\text{m/s}^2\) into the equation to find the speed:\[ v = \sqrt{2 \times 9.8 \times (10 \sqrt{2} - 10 \sqrt{3})}\]Simplify the inside of the square root and calculate \(v\):\[ v = \sqrt{19.6 \times (10 \sqrt{2} - 10 \sqrt{3})}\]Calculate the numeric value to find the speed in m/s.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

trigonometry in physics
Trigonometry plays a crucial role in physics, especially when dealing with motion and forces at various angles. When Tarzan swings from one tree to another, we can use trigonometric functions to understand his motion at different points. The vine he swings from forms angles of 45掳 and 30掳 with the vertical at the start and end of his journey, respectively.
To find the vertical height that Tarzan reaches, we use the cosine function. For a triangle where the hypotenuse is the vine (20 meters), the height is found by:
  • Start Height: \( h_i = 20 \cos(45掳) \).
  • End Height: \( h_f = 20 \cos(30掳) \).
These calculations help us understand how high Tarzan is above the lowest point of his swing at each stage, ultimately influencing his speed before reaching Jane.
gravitational potential energy
Gravitational potential energy (GPE) is the energy stored in an object as a result of its vertical position or height. As Tarzan swings, he moves through different heights, converting potential energy to kinetic energy. His initial and final potential energies are based on these heights, found using trigonometry.
We calculate the change in height \( \Delta h \) as:
  • Initial height from 45掳: \( h_i = 10 \sqrt{2} \)
  • Final height from 30掳: \( h_f = 10\sqrt{3} \)
  • Change in height: \( \Delta h = 10 \sqrt{2} - 10 \sqrt{3} \)
With the change in height, we determine the change in GPE using the formula \( \Delta PE = mg\Delta h \), where \( m \) is Tarzan's mass and \( g = 9.8 \, \mathrm{m/s^2} \). This energy transformation is key to understanding how Tarzan moves smoothly and speedily towards Jane.
kinetic energy
Kinetic energy (KE) is the energy of motion. In Tarzan's swing, gravitational potential energy is converted into kinetic energy, propelling him forward. The formula for kinetic energy is \( KE = \frac{1}{2}mv^2 \), where \( v \) is Tarzan's speed. At the start, Tarzan has potential energy due to his height, but as he descends, this energy converts to kinetic energy.
We assume no energy loss in the system, so the change in potential energy becomes kinetic energy:
  • \( \Delta PE = \Delta KE = \frac{1}{2}mv^2 \)
To find Tarzan's speed just before reaching Jane, we solve for \( v \):
  • Rearranging gives \( v^2 = 2g (10 \sqrt{2} - 10 \sqrt{3}) \)
  • Substitute \( g = 9.8 \, \mathrm{m/s^2} \) and solve \( v = \sqrt{\text{calculated value}} \)
This calculation shows how Tarzan's initial energy is transformed into motion. Understanding this conversion is a beautiful demonstration of the conservation of energy in physics.

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Most popular questions from this chapter

A baseball is thrown from the roof of a 22.0 - -tall building with an initial velocity of magnitude 12.0 \(\mathrm{m} / \mathrm{s}\) and directed at an angle of \(53.1^{\circ}\) above the horizontal. (a) What is the speed of the ball just before it strikes the ground? Use energy methods and ignore air resistance. (b) What is the answer for part (a) if the initial velocity is at an angle of \(53.1^{\circ}\) below the horizontal? \((\mathrm{c})\) If the effects of air resistance are are included, will part (a) or (b) give the higher speed?

A 2.00 -kg block is pushed against a spring with negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m},\) compressing it 0.220 \(\mathrm{m}\) . When the block is released, it moves along a frictionless, horizontal surface and then up a frictionless incline with slope 37.0" (Fig. \(\mathrm{P7} .42\) ). (a) What is the speed of the block as it slides along the horizontal surface after having left the spring? (b) How far does the block travel up the incline before starting to slide back down?

If a fish is attached to a vertical spring and slowly lowered to its equilibrium position, it is found to stretch the spring by an amount \(d .\) If the same fish is attached to the end of the unstretched spring and then allowed to fall from rest, through what maximum distance does it stretch the spring? (Hint: Calculate the force constant of the spring in terms of the distance \(d\) and the mass \(m\) of the fish.)

A hydroelectric dam holds back a lake of surface area \(3.0 \times 10^{6} \mathrm{m}^{2}\) that has vertical sides below the water level. The water level in the lake is 150 \(\mathrm{m}\) above the base of the dam. When the water passes through turbines at the base of the dam, its mechanical energy is converted to electrical energy with 90\(\%\) efficiency. (a) If gravitational potential energy is taken to be zero at the base of the dam, how much energy is stored in the top meter of the water in the lake? The density of water is 1000 \(\mathrm{kg} / \mathrm{m}^{3}\) . (b) What volume of water must pass through the dam to produce 1000 kilowatt-hours of electrical energy? What distance does the level of water in the lake fall when this much water passes through the dam?

The Great Sandini is a 60 -kg circus performer who is shot from a cannon (actually a spring gun). You don't find many men of his caliber, so you help him design a new gun. This new gun has a very large spring with a very small mass and a force constant of 1100 \(\mathrm{N} / \mathrm{m} / \mathrm{m}\) that he will compress with a force of 4400 \(\mathrm{N}\) .The inside of the gun barrel is coated with Teflon, so the average friction force will be only 40 \(\mathrm{N}\) during the 4.0 \(\mathrm{m}\) he moves in the barrel. At what speed will he emerge from the end of the barrel, 2.5 \(\mathrm{m}\) above his initial rest position?
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