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Suppose an Olympic diver who weighs \(52.0 \mathrm{~kg}\) executes a straight dive from a 10 -m platform. At the apex of the dive, the diver is \(10.8 \mathrm{~m}\) above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) Assuming that all the potential energy of the diver is converted into kinetic energy at the surface of the water, at what speed, in \(\mathrm{m} / \mathrm{s}\), will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

Short Answer

Expert verified
The potential energy of the diver at the apex of the dive is approximately 5529.216 Joules. The diver will enter the water at a speed of approximately 14.27 m/s. Yes, the diver does work on entering the water, as they transfer energy to the water through the force they apply while entering.

Step by step solution

01

Part (a): Calculate the potential energy at the apex of the dive

To calculate the potential energy (PE) of the diver at the peak of the dive, we can use the potential energy formula: \(PE = mgh\) Where: \(m = 52.0\text{ kg}\) (mass of the diver) \(g = 9.81 \frac{\text{m}}{\text{s}^2}\) (acceleration due to gravity) \(h = 10.8 \text{ m}\) (height above the water surface) Plugging in the values we have: \(PE = (52.0\text{ kg})(9.81\frac{\text{m}}{\text{s}^2})(10.8\text{ m})\)
02

Part (a): Compute the potential energy

Now, we can compute the potential energy of the diver at the peak of the dive: \(PE = 5529.216\text{ J}\) So, the potential energy of the diver at the apex of the dive, relative to the surface of the water, is approximately 5529.216 Joules.
03

Part (b): Use conservation of mechanical energy

We can assume the mechanical energy of the diver is conserved during the dive. This means that the potential energy at the highest point is all converted into kinetic energy (KE) just before entering the water. Thus, \(PE = KE\) We can use the kinetic energy formula to find the speed of the diver at the water surface: \(KE = \frac{1}{2}mv^2\) Where: \(v\) is the speed of the diver, and the other variables have been defined earlier.
04

Part (b): Calculate the speed at the surface of the water

Now, we can solve for the speed (\(v\)) of the diver just before entering the water: \(5529.216\text{ J} = \frac{1}{2}(52.0\text{ kg})v^2\) Now we find v: \(v = \sqrt{\frac{2 \times 5529.216\text{ J}}{52.0\text{ kg}}}\)
05

Part (b): Compute the dive speed

Finally, we can compute the speed of the diver as they enter the water: \(v \approx 14.27\frac{\text{m}}{\text{s}}\) So, the diver will enter the water at a speed of approximately 14.27 m/s.
06

Part (c): Does the diver do work on entering the water?

When the diver enters the water, the water applies a force on the diver to slow them down. This force is in the opposite direction of the diver's motion, so it does negative work on the diver. In turn, the diver applies force on the water as they enter it (Newton's third law), which means the diver does work on the water. This work done by the diver on the water results in the transfer of energy from the diver to the water (as some of the diver's kinetic energy is lost to heat, sound, and water displacement). So, yes, the diver does work on entering the water.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a concept that helps explain the movement of objects. Whenever an object is in motion, it possesses kinetic energy. This form of energy depends on two primary factors: the mass of the object and its speed. The mathematical expression for kinetic energy is given by the formula \[ KE = \frac{1}{2}mv^2 \]Here, \( KE \) represents the kinetic energy, \( m \) is the mass of the object, and \( v \) is its velocity. It shows that kinetic energy increases with the square of an object's speed, meaning that an object moving twice as fast will have four times the kinetic energy. When analyzing problems involving motion, such as the Olympic diver in our exercise, it's essential to think about how potential energy turns into kinetic energy as the diver drops from the 10-m platform to the water's surface. At the highest point, the diver's potential energy is at its peak, and their kinetic energy is zero. As the diver descends, potential energy is gradually converted into kinetic energy. Just before they enter the water, their potential energy is zero, while their kinetic energy is at its maximum.
Conservation of Mechanical Energy
The principle of conservation of mechanical energy states that if no external forces (like friction or air resistance) are doing work, the total mechanical energy (potential + kinetic) of a system remains constant. This principle is crucial in solving problems related to motion and energy. In the case of our Olympic diver, this principle allows us to assume that all of the potential energy at the apex of the dive will convert to kinetic energy by the time the diver reaches the water's surface. Here's why it matters:
  • The potential energy of an object at a height \( h \) is given by \( mgh \).
  • Once the diver leaps off, their potential energy decreases as it is converted to kinetic energy.
  • According to the conservation of mechanical energy, just before reaching the water, the diver's kinetic energy is equal to the initial potential energy at the peak.
This interchange between potential and kinetic energy enables us to predict the speed of the diver upon entering the water accurately. It simplifies calculations and deepens our understanding of how energy transformations occur in a closed system without losing energy to outside forces.
Work and Energy Principles
The work and energy principles link the concepts of work and energy, showing how they interact to influence an object's motion. Work is done when a force causes an object to move, and it can either add or remove energy from a system. The amount of work done is calculated as:\[ W = Fd \cos \theta \]Here \( W \) represents work, \( F \) is the applied force, \( d \) is the displacement, and \( \theta \) is the angle between the force and displacement directions.In the exercise scenario, when the diver enters the water, the water exerts a force against the diver's downward motion, doing negative work on the diver. Since the force opposes the motion, it decreases the diver's kinetic energy and eventually brings the diver to a stop. Moreover, the diver also exerts force on the water, displacing it. Newton's third law states that for every action, there is an equal and opposite reaction. Consequently, the diver does work on the water, transferring energy and causing water displacement and splashes. As energies convert and transfer during this interaction, work and energy principles make it easier to see how energy leaves the diver and enters surrounding systems like sound and heat.

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Most popular questions from this chapter

A \(100-\mathrm{kg}\) man decides to add to his exercise routine by walking up six flights of stairs \((30 \mathrm{~m}) 10\) times per day. He figures that the work required to increase his potential energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption?

A sample of a hydrocarbon is combusted completely in \(\mathrm{O}_{2}(g)\) to produce \(21.83 \mathrm{~g} \mathrm{CO}_{2}(g), 4.47 \mathrm{~g} \mathrm{H}_{2} \mathrm{O}(g),\) and \(311 \mathrm{~kJ}\) of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of \(\Delta H_{f}^{\circ}\) per empiricalformula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer.

In a thermodynamic study, a scientist focuses on the properties of a solution in an apparatus as illustrated. A solution is continuously flowing into the apparatus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolated system? (b) If the inlet and outlet were closed, what type of system would it be?

The gas-phase reaction shown, between \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\), was run in an apparatus designed to maintain a constant pressure. (a) Write a balanced chemical equation for the reaction depicted and predict whether \(w\) is positive, negative, or zero. (b) Using data from Appendix C, determine \(\Delta H\) for the formation of one mole of the product. [Sections 5.3 and 5.7\(]\)

The hydrocarbons cyclohexane \(\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-156\right.\) \(\mathrm{kJ} / \mathrm{mol}\) ) and 1-hexene \(\left(\mathrm{C}_{6} \mathrm{H}_{12}(l), \Delta H_{f}^{\circ}=-74 \mathrm{~kJ} / \mathrm{mol}\right)\) have the same empirical formula. (a) Calculate the standard enthalpy change for the transformation of cyclohexane to 1-hexene. (b) Which has greater enthalpy, cyclohexane or 1-hexene? (c) Without doing a further calculation and knowing the answer to (b), do you expect cyclohexane or 1-hexene to have the larger combustion enthalpy?

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