Question

Tarzan swings on a taut vine from his tree house to a limb on a neighboring tree, which is located a horizontal distance of \(10.0 \mathrm{~m}\) from and a vertical distance of \(4.00 \mathrm{~m}\) below his starting point. Amazingly the vine neither stretches nor breaks; Tarzan's trajectory is thus a portion of a circle. If Tarzan starts with zero speed, what is his speed when he reaches the limb?

Short answer

Answer: Tarzan's final speed when he reaches the limb is 8.85 m/s.

Step-by-step solution

Identify the information given

We are given the following information: - Horizontal distance from the tree house to the limb: \(d_x = 10.0 \, \mathrm{m}\) - Vertical distance from the tree house to the limb: \(d_y = 4.00 \, \mathrm{m}\) - Tarzan starts with zero speed

Calculate change in gravitational potential energy

First, we need to find the change in gravitational potential energy between Tarzan's initial and final positions. This can be calculated using the formula: \(ΔU = mgh\) where: - \(ΔU\) is the change in potential energy - \(m\) is Tarzan's mass - \(g\) is Earth's gravitational acceleration \((9.81 \, \mathrm{m/s^2})\) - \(h\) is the height difference between Tarzan's initial and final positions \(h = 4.00 \, \mathrm{m}\) Now, we can calculate the change in gravitational potential energy: \(ΔU = mgh = mg(4 \, \mathrm{m})\)

Use conservation of energy

Since energy is conserved, the kinetic energy gained by Tarzan will be equal to the change in gravitational potential energy. Therefore: \(ΔK = ΔU\) The formula for kinetic energy is: \(K = \frac{1}{2}mv^2\) where \(v\) is the speed of Tarzan. Since Tarzan starts with zero speed, his initial kinetic energy \(K_1\) is zero. Therefore, the change in kinetic energy will be equal to the final kinetic energy \(K_2\). We can write the conservation of energy equation as follows: \(K_2 = ΔU\)

Solve for Tarzan's final speed

Using the conservation of energy equation, we can solve for Tarzan's final speed, \(v\): \(\frac{1}{2}mv^2 = mg(4 \, \mathrm{m})\) Now, we can cancel out the mass \(m\) from both sides and solve for \(v\): \(\frac{1}{2}v^2 = g(4 \, \mathrm{m})\) Multiply both sides by 2: \(v^2 = 2g(4 \, \mathrm{m})\) Finally, take the square root of both sides: \(v = \sqrt{2g(4 \, \mathrm{m})} = \sqrt{2(9.81 \, \mathrm{m/s^2})(4 \, \mathrm{m})} = 8.85 \, \mathrm{m/s}\) So, Tarzan's speed when he reaches the limb is \(8.85 \, \mathrm{m/s}\).

Key concepts

Gravitational Potential Energy

Gravitational potential energy (GPE) is the energy an object possesses due to its position in a gravitational field, most commonly the gravitational field of the Earth. An easy way to understand GPE is to think of it as stored energy that has the potential to do work. When Tarzan swings down from a higher point to a lower one, he is moving in the Earth's gravitational field.

GPE is directly proportional to three factors: the object's mass (\(m\)), the height (\(h\)) above the reference point, often the ground, and the gravitational acceleration (\(g\)) on Earth, which is approximately \(9.81 \text{ m/s}^2\text{. The formula to calculate it is \(GPE = mgh\text{. In the context of our exercise, Tarzan starts with a certain amount of GPE at his initial height, which is then converted into kinetic energy as he descends.

Kinetic Energy

Kinetic energy (KE) is the energy of motion. Any object that is in motion, whether it's a car driving down the street or an apple falling from a tree, has kinetic energy. The formula to calculate the kinetic energy of an object is \(KE = \frac{1}{2}mv^2\text{, where \(m\) is mass and \(v\) is the velocity (or speed) of the object.

In Tarzan's scenario, as he swings and descends, the gravitational potential energy is converted into kinetic energy. Since no other forces (like air resistance) are considered in the textbook problem, all the initial GPE is converted into KE. This is a demonstration of the conservation of energy principle: the total energy remains constant if no external forces are doing work. Tarzan's speed at the limb can be found by equating the change in GPE to the final KE.

Projectile Motion

Connecting Projectile Motion with Energy

While the exercise does not explicitly describe Tarzan's motion as projectile motion, this concept is closely related. Projectile motion refers to the motion of an object thrown or projected into the air, subject only to acceleration as a result of gravity. Common examples include a football thrown in a parabolic arc or a stone launched by a slingshot.

In Tarzan's case, his pendulous swing through the air is a portion of circular motion rather than traditional projectile motion. However, understanding projectile motion is helpful because it involves both GPE and KE. At the highest point in projectile motion, an object has maximum GPE and minimum KE. As the object falls, its GPE is converted into KE, causing it to speed up if we ignore air resistance.

Projectile motion typically has two components: horizontal and vertical. The horizontal motion occurs at a constant speed (if air resistance is negligible), and the vertical motion is affected by gravity, just like Tarzan's swing. By analyzing both components, one can understand the overall trajectory of the object—although Tarzan's circular trajectory involves more complex forces and energy conversions.