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Chapter 7: Problem 12

Tarzan, in one tree, sights Jane in another tree. He grabs the end of a vine with length 20 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. Ignore air resistance and the mass of the vine.

Short Answer

Expert verified
Tarzan does not reach Jane as the calculated speed does not allow for a tender embrace.

Step by step solution

01

Determine the Height Difference

First, we need to calculate the difference in height between the initial and final positions of Tarzan to determine from what height he falls. When the vine is at an angle of 45掳 with the vertical, the vertical component of the length of the vine can be calculated as \( h_1 = L \cdot \cos(45^\circ) \). The same applies when the vine is at an angle of 30掳 with the vertical, yielding \( h_2 = L \cdot \cos(30^\circ) \). Therefore the height difference is \( \Delta h = h_1 - h_2 \).
02

Initial and Final Height Calculation

Given that \( L = 20 \) m, we find the heights: - Initial height: \( h_1 = 20 \cdot \cos(45^\circ) = 20 \cdot \frac{\sqrt{2}}{2} = 10\sqrt{2} \) m. - Final height: \( h_2 = 20 \cdot \cos(30^\circ) = 20 \cdot \frac{\sqrt{3}}{2} = 10\sqrt{3} \) m. Therefore, \( \Delta h = 10\sqrt{2} - 10\sqrt{3} \).
03

Calculate Tarzan's Speed Using Conservation of Energy

Using the conservation of energy principle, we know the potential energy lost is converted to kinetic energy. The potential energy change is \( \Delta PE = mg\Delta h \), and the kinetic energy gained is \( KE = \frac{1}{2}mv^2 \).Setting \( \Delta PE = KE \), we have:\[ mg\Delta h = \frac{1}{2}mv^2 \]Hence, solving for speed \( v \):\[ v = \sqrt{2g\Delta h} \]
04

Substitute Values and Compute

Substituting \( g = 9.8 \text{ m/s}^2 \) and \( \Delta h = 10(\sqrt{2} - \sqrt{3}) \), we calculate: \[ v = \sqrt{2 \times 9.8 \times 10(\sqrt{2} - \sqrt{3})} \approx \sqrt{-19.6 \times (\sqrt{2} - \sqrt{3})} \approx \sqrt{-60.48} \].This leads us to the conclusion that we have a negative value, indicating Tarzan does not have enough height to gain kinetic energy to reach Jane.

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Key Concepts

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

Conservation of Energy
In physics, the conservation of energy is a vital principle. It states that the total energy in an isolated system remains constant. This means energy can't be created or destroyed, only transformed.
For Tarzan's swing, the potential energy he has at the start, when he is high up in the tree, is converted into kinetic energy as he swings towards Jane.
  • Potential energy depends on his height relative to the ground.
  • Kinetic energy depends on his speed.
By applying the conservation of energy, you can predict how fast Tarzan will be moving when he reaches Jane if no energy were to be lost to other forces like air resistance. In this problem, Tarzan's initial potential energy becomes the kinetic energy necessary for reaching maximum speed at the lowest point of the swing.
Kinematics
Kinematics is the branch of physics that deals with the motion of objects without considering the causes of motion. It involves concepts such as displacement, velocity, and acceleration. For Tarzan's swing, understanding these concepts helps predict his motion through space.
  • Velocity indicates how fast an object changes its position.
  • Acceleration shows how velocity changes over time.
In the exercise, Tarzan converts his initial static start on the vine into motion, where we explore how fast he is going when he reaches Jane. Though kinematics doesn't delve into the forces causing the motion, it helps calculate the motion through parameters like angle and length of the vine, and height from the initial to final positions.
Angular Motion
When Tarzan swings on a vine, his motion can be described as angular motion. This is because the vine acts as a pivot similar to a pendulum, and he moves in a curving path.
Angular motion connects directly to the linear motion Tarzan undergoes on his way to Jane.
  • The angle of the vine from the vertical at the start and end are key.
  • His displacement through the arc of the vine swing is determined by changes in angle.
Understanding angular motion helps predict how displacement and velocity change during the swing. Concepts like angular velocity and angular displacement become part of the analysis when evaluating how Tarzan's speed varies as he moves along a path defined by an arc.
Potential Energy
Potential energy is the energy held by an object because of its position relative to other objects. In Tarzan's case, it is due to his height above the ground at the start of his swing.
  • The higher Tarzan is, the more potential energy he has.
  • Potential energy can be calculated using the formula: \( PE = mgh \), where \( m \) is mass, \( g \) is gravitational force, and \( h \) is height from the ground.
Initially, Tarzan has maximum potential energy when he is higher in the tree. This energy is gradually converted into kinetic energy as he descends, allowing him to accelerate through his swing. When he calculates the difference in height, it provides the amount of potential energy available for conversion to movement.
Kinetic Energy
Kinetic energy is the energy of motion, which depends on an object's mass and velocity. For Tarzan, it's about how fast he is moving just before reaching Jane.

The kinetic energy formula is \( KE = \frac{1}{2}mv^2 \). This tells us that any increase in velocity, as Tarzan swings down, results in a greater increase in kinetic energy than an equal increase in mass would.
  • At the lowest part of the swing, potential energy is minimal, and kinetic energy maximizes.
  • In Tarzan's problem, kinetic energy helps determine whether his embrace with Jane will be gentle or forceful.
By analyzing the conversions between potential and kinetic energy, we get insights into Tarzan鈥檚 possible speed at various points of his movement, crucial for defining his final speed before he meets Jane.

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Most popular questions from this chapter

A small block with mass 0.0400 kg is moving in the \(xy\)-plane. The net force on the block is described by the potentialenergy function \(U(x, y) = (5.80 \, \mathrm{J/m}^2)x^2 - (3.60 \, \mathrm{J/m}^3)y^3\). What are the magnitude and direction of the acceleration of the block when it is at the point (\(x =\) 0.300 m, \(y =\) 0.600 m)?

A 2.50-kg mass is pushed against a horizontal spring of force constant 25.0 N/cm on a frictionless air table. The spring is attached to the tabletop, and the mass is not attached to the spring in any way. When the spring has been compressed enough to store 11.5 J of potential energy in it, the mass is suddenly released from rest. (a) Find the greatest speed the mass reaches. When does this occur? (b) What is the greatest acceleration of the mass, and when does it occur?

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.)

In an experiment, one of the forces exerted on a proton is \(\overrightarrow{F}\) \(= -a x^2 \hat{\imath}\), where \(\alpha = 12 \mathrm{N/m}^2\). (a) How much work does \(\overrightarrow{F}\) do when the proton moves along the straight-line path from the point (0.10 m, 0) to the point (0.10 m, 0.40 m)? (b) Along the straight-line path from the point (0.10 m, 0) to the point (0.30 m, 0)? (c) Along the straight-line path from the point (0.30 m, 0) to the point (0.10 m, 0)? (d) Is the force \(\overrightarrow{F}\) conservative? Explain. If \(\overrightarrow{F}\) is conservative, what is the potential-energy function for it? Let \(U =\) 0 when \(x =\) 0.

A spring of negligible mass has force constant \(k =\) 800 N/m. (a) How far must the spring be compressed for 1.20 J of potential energy to be stored in it? (b) You place the spring vertically with one end on the floor. You then lay a 1.60-kg book on top of the spring and release the book from rest. Find the maximum distance the spring will be compressed.
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