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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. Ignore air resistance and the mass of the vine.

Short Answer

Expert verified
Tarzan's speed, just before he reaches Jane, is approximately \(8.5 \mathrm{m/s}\). Whether he tenderly embraces her or knocks her off, however, may depend on many other factors like their masses and reactions.

Step by step solution

01

Identify the initial and final positions

Tarzan starts at the top of one swing where his velocity is zero. His kinetic energy is all potential due to his height, \(h1\), above the lowest point of the swing. In the final position, he's at a height, \(h2\), above the lowest point. Here, his kinetic energy has both potential and kinetic components.
02

Establish height of Tarzan from the lowest point

First, the full length of the vine needs to be determined. As the vine initially makes an angle of \(45^{\circ}\) with the vertical, using a bit of trigonometry, the height, \(h1\), from the lowest point to the starting position can be calculated as \(20(1-\cos(45^{\circ})) = 20(1-\frac{\sqrt{2}}{2}) \approx 5.86 \mathrm{~m}\). Similarly, when he arrives at Jane's position, the vine makes an angle of \(30^{\circ}\) with the vertical, so \(h2 = 20(1-\cos(30^{\circ})) = 20(1-\frac{\sqrt{3}}{2}) \approx 2.32 \mathrm{~m}\).
03

Calculate Tarzan's final speed

Using the principle of conservation of mechanical energy (since no non-conservative forces like air resistance or friction are doing work), the sum of kinetic and potential energy at the two positions (initial and final) are equal. Therefore, \(mgh1 = mgh2 + \frac{1}{2} mv^2\), where \(m\) is Tarzan's mass, \(g\) is the acceleration due to gravity, and \(v\) is his speed. Solving for \(v\), we get \(v = \sqrt{2g(h1-h2)}\). We don't know Tarzan's mass, but it cancels out in our equation. Substituting for \(h1\), \(h2\) and \(g = 9.8 \mathrm{m/s}^2\), we find \(v \approx \sqrt{2*9.8*(5.86-2.32)} \approx 8.5 \mathrm{m/s}\),his speed just before reaching Jane.

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Key Concepts

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

Conservation of Mechanical Energy
Understanding the conservation of mechanical energy is crucial in solving problems related to the motion of objects where only conservative forces, like gravity, are at work. Mechanical energy consists of two primary forms: kinetic energy, the energy of motion, and potential energy, the stored energy an object has due to its position or configuration. When no non-conservative forces (such as friction or air resistance) are acting on an object, its total mechanical energy remains constant throughout its motion.

This principle allows us to analyze perplexing situations, like Tarzan swinging through the jungle, without knowing specific details such as his mass, simply based on his position and speed at different points of his trajectory. By setting the total mechanical energy at the start equal to the total at any later point, we can solve for unknowns — like Tarzan's speed before he reaches Jane. The clever canceling out of certain variables, like mass, further simplifies the computations, making this principle not just mighty, but incredibly versatile in physics problem-solving.
Kinetic and Potential Energy
Let's dive a bit deeper into the specific forms of energy discussed under the principle of conservation of mechanical energy. Kinetic energy is the result of an object's motion and is defined by the equation \(\frac{1}{2}mv^2\), where \(m\) is mass and \(v\) is velocity. This form of energy is always a positive value since it's proportional to the square of the velocity.

Potential Energy

In contrast, potential energy is associated with an object's position or configuration. For our swinging Tarzan, the gravitational potential energy, given by \(mgh\), is most relevant. It varies with the height \(h\) above a reference level and the mass \(m\) within the gravitational field of strength \(g\), the acceleration due to gravity. As Tarzan swings higher, his kinetic energy is converted into potential energy and vice versa. This interplay ensures that the overall mechanical energy, considering isolated systems, is conserved, leading to a predictable and calculable outcome.
Trigonometry in Physics
Trigonometry, the branch of mathematics dealing with the relationships between the angles and sides of triangles, is not just for abstract theory; it's a powerful tool in physics, especially when analyzing forces and motion in various dimensions. In our Tarzan scenario, trigonometry helps determine the heights at different points of the swing by relating the length of the vine to the angles it makes with the vertical.

Using the cosine function, which compares the adjacent side of a right-angled triangle to the hypotenuse, we calculate the vertical component of the vine's length. This gives us the initial and final heights of Tarzan above the ground. It may seem like a simple calculation, but understanding these relationships is key for physics problem-solving. Trigonometry bridges the gap between abstract angles and tangible measures of distance, which helps students like you transform angled trajectories into solvable physics puzzles.

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Most popular questions from this chapter

\(A 2.50 \mathrm{~kg}\) block on a horizontal floor is attached to a horizontal spring that is initially compressed \(0.0300 \mathrm{~m}\). The spring has force constant \(840 \mathrm{~N} / \mathrm{m}\). The coefficient of kinetic friction between the floor and the block is \(\mu_{k}=0.40 .\) The block and spring are released from rest, and the block slides along the floor. What is the speed of the block when it has moved a distance of \(0.0200 \mathrm{~m}\) from its initial position? (At this point the spring is compressed \(0.0100 \mathrm{~m}\).)

A wooden block with mass \(1.50 \mathrm{~kg}\) is placed against a compressed spring at the bottom of an incline of slope \(30.0^{\circ}\) (point \(A\) ). When the spring is released, it projects the block up the incline. At point \(B,\) a distance of \(6.00 \mathrm{~m}\) up the incline from \(A\), the block is moving up the incline at \(7.00 \mathrm{~m} / \mathrm{s}\) and is no longer in contact with the spring. The coefficient of kinetic friction between the block and the incline is \(\mu_{k}=0.50\) The mass of the spring is negligible. Calculate the amount of potential energy that was initially stored in the spring.

A 90.0 kg mail bag hangs by a vertical rope 3.5 m long. A postal worker then displaces the bag to a position 2.0 m sideways from its original position, always keeping the rope taut. (a) What horizontal force is necessary to hold the bag in the new position? (b) As the bag is moved to this position, how much work is done (i) by the rope and (ii) by the worker?

A 25.0 kg child plays on a swing having support ropes that are 2.20 m long. Her brother pulls her back until the ropes are 42.0° from the vertical and releases her from rest. (a) What is her potential energy just as she is released, compared with the potential energy at the bottom of the swing’s motion? (b) How fast will she be moving at the bottom? (c) How much work does the tension in the ropes do as she swings from the initial position to the bottom of the motion?

A \(2.00 \mathrm{~kg}\) 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 \(37.0^{\circ}\) (Fig. \(\mathbf{P 7 . 4 0}\) ). (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?

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