Problem 43 - College Physics

Question

\(\bullet\) The total height of Yosemite Falls is 2425 ft. (a) How many more joules of gravitational potential energy are there for each kilogram of water at the top of this waterfall com- pared with each kilogram of water at the foot of the falls? (b) Find the kinetic energy and speed of each kilogram of water as it reaches the base of the waterfall, assuming that there are no losses due to friction with the air or rocks and that the mass of water had negligible vertical speed at the top. How fast (in \(\mathrm{m} / \mathrm{s}\) and mph) would a 70 \(\mathrm{kg}\) person have to run to have that much kinetic energy? (c) How high would Yosemite Falls have to be so that each kilogram of water at the base had twice the kinetic energy you found in part (b); twice the speed you found in part (b)?

Step-by-Step Solution

Answer

(a) 7249.45 J. (b) 120.41 m/s or 269.33 mph (for 70 kg, 14.39 m/s or 32.20 mph). (c) 1478.28 m for double KE, 2956.54 m for double speed.

Step 1: Convert height to meters

The height of Yosemite Falls is 2425 feet. To find the energy difference, we need to convert this height into meters, as the gravitational potential energy formula uses meters. We use the conversion factor: 1 foot = 0.3048 meters. Thus, \( h = 2425 \times 0.3048 = 739.14 \text{ meters} \).

Step 2: Calculate gravitational potential energy per kilogram

The gravitational potential energy (GPE) for an object at height \( h \) is given by \( PE = mgh \), where \( m = 1 \text{ kg} \) (mass per kg), \( g \approx 9.81 \text{ m/s}^2 \) (acceleration due to gravity), and \( h = 739.14 \text{ m} \). Thus, \( PE = 1 \times 9.81 \times 739.14 = 7249.45 \text{ joules} \).

Step 3: Find kinetic energy at the falls' base

Assuming no energy losses, all gravitational potential energy converts to kinetic energy at the base. Therefore, the kinetic energy (KE) each kilogram has is \( KE = 7249.45 \text{ joules} \).

Step 4: Calculate speed from kinetic energy

The kinetic energy formula is \( KE = \frac{1}{2} mv^2 \). Solving for velocity \( v \), with \( KE = 7249.45 \text{ joules} \) and \( m = 1 \text{ kg} \), yields \( 7249.45 = \frac{1}{2} \times 1 \times v^2 \). Thus, \( v^2 = 14498.9 \), so \( v = \sqrt{14498.9} = 120.41 \text{ m/s} \).

Step 5: Convert speed to mph

To convert 120.41 m/s to mph, use the factor: 1 m/s = 2.23694 mph. Thus, \( 120.41 \times 2.23694 \approx 269.33 \text{ mph} \).

Step 6: Calculate running speed for 70 kg person

For a 70 kg person to have 7249.45 J of KE, use \( KE = \frac{1}{2} m v^2 \). \( v^2 = \frac{2 \times 7249.45}{70} \), giving \( v^2 = 207.127 \). Thus, \( v = \sqrt{207.127} = 14.39 \text{ m/s} \), or using the conversion, approximately 32.20 mph.

Step 7: Calculate new height for double kinetic energy

To double the kinetic energy, \( KE = 2 \times 7249.45 = 14498.9 \text{ joules} \). \( mgh = 14498.9 \) leads to \( h = \frac{14498.9}{9.81} \approx 1478.28 \text{ m}\).

Step 8: Calculate new height for double speed

Doubling speed means doubling kinetic energy in terms of the squared function: if \( v \) doubles, KE becomes four times. So, \( 4 \times 7249.45 = 28997.8 \text{ joules} \). Solving \( mgh = 28997.8 \) gives \( h = \frac{28997.8}{9.81} \approx 2956.54 \text{ m} \).

Key Concepts

Kinetic Energy

Kinetic energy refers to the energy possessed by an object due to its motion. When an object is in motion, it has the ability to perform work due to its kinetic energy. This energy is quantified using the equation \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity. In the context of a waterfall, as water descends and its speed increases, it gains kinetic energy, showcasing energy conversion from gravitational potential energy to kinetic energy. Understanding this concept helps to determine the speed and impact force of descending water.

Energy Conversion

Energy conversion plays a crucial role in physics, illustrating how energy changes from one form to another. In our scenario, the gravitational potential energy of water at the top of Yosemite Falls is converted into kinetic energy as water falls. This conversion is ideally free of energy loss when considering negligible friction or air resistance. Thus, the gravitational energy calculated by \( PE = mgh \) at the waterfall's top equals the kinetic energy \( KE = \frac{1}{2} mv^2 \) at the base. This energy transformation emphasizes the conservation of energy principle, which states that total energy remains constant, merely changing forms.

Speed Calculation

Speed calculation from kinetic energy is essential to understand how quickly an object moves. With kinetic energy known, the relationship \( KE = \frac{1}{2} mv^2 \) lets us find the speed \( v \). Solving for \( v \), this becomes \( v = \sqrt{\frac{2 \, KE}{m}} \). In the Yosemite Falls example, we calculate the speed at the bottom using the kinetic energy derived from the waterfall's height. This calculation shows how elevation influences speed. It also applies to converting meters per second to miles per hour for contextual understanding.

Yosemite Falls Physics

Yosemite Falls provides a fascinating case study in physics due to its significant height. Its towering drop allows for robust analysis of gravitational potential energy and energy conversion. At 2,425 feet, the potential energy is considerable, resulting in significant kinetic energy and speed at the base. This exercise illustrates textbook physics principles, emphasizing the predictability and calculations involved when analyzing natural phenomena like waterfalls. By assessing energy conversion at Yosemite Falls, students better understand real-world applications of physics principles.

Height Conversion

Height conversion is critical when solving physics problems, especially when dealing with gravitational potential energy calculations, which require consistency in units. In this exercise, we convert the height of Yosemite Falls from feet to meters, using the conversion factor 1 foot = 0.3048 meters. This ensures gravitational potential energy is calculated correctly, as it depends on the height in meters. Proper unit conversion allows for accurate analysis of the gravitational force and potential energy changes. It's a fundamental step for precise and meaningful results in physics calculations.

Metric Conversions in Physics

Metric conversions are essential in physics to maintain uniform units across calculations, ensuring accuracy and simplicity. Converting between feet, meters, and speed units like meters per second and miles per hour avoids errors. Physics relies on the International System of Units (SI), which often means converting customary units like feet to meters. The practice of metric conversions aids clear communication and understanding of physical quantities and their relationships. Mastery of these conversions is a foundational skill that supports more complex problem-solving in physics.