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Chapter 7: Problem 32

\(\bullet\) A 72.0 -kg swimmer jumps into the old swimming hole from a diving board 3.25 m above the water. Use energy conserva- tion to find his speed just he hits the water (a) if he just holds his nose and drops in, (b) if he bravely jumps straight up (but just beyond the board!) at \(2.50 \mathrm{m} / \mathrm{s},\) and \((\mathrm{c})\) if he manages to jump downward at 2.50 \(\mathrm{m} / \mathrm{s}\) .

Short Answer

Expert verified
(a) 7.98 m/s, (b) 8.33 m/s, (c) 8.59 m/s.

Step by step solution

01

Understanding Energy Conservation

We start by identifying that the swimmer converts potential energy (due to height) into kinetic energy (due to speed) as they fall. The total mechanical energy is conserved (if we neglect air resistance). The key equation here is \( mgh = \frac{1}{2}mv^2 \).
02

Calculate Potential Energy at the Start

Potential energy at the start is given by \( PE = mgh \), where \( m \) is the mass (72.0 kg), \( g \) is the acceleration due to gravity (9.8 m/s虏), and \( h \) is the height (3.25 m). Substitute to find \( PE = 72 \times 9.8 \times 3.25 \).
03

Calculate Speed for Part (a): Just Holds His Nose and Drops

For part (a), the swimmer just drops, so initial kinetic energy is zero. Set potential energy equal to kinetic energy: \( mgh = \frac{1}{2}mv^2 \). Solve for \( v \): \( v = \sqrt{2gh} \). Substitute the values: \( v = \sqrt{2 \times 9.8 \times 3.25} \) to find the speed.
04

Calculate Speed for Part (b): Jumps Straight Up

For part (b), the swimmer jumps up with an initial speed. Use \( K.E._{initial} = \frac{1}{2}mv_0^2 \) (\( v_0 = 2.5 \mathrm{m/s} \)): \( v^2 = v_0^2 + 2gh \). Substitute to calculate: \( v^2 = (2.50)^2 + 2 \times 9.8 \times 3.25 \) and solve for \( v \).
05

Calculate Speed for Part (c): Jumps Downward

For part (c), the swimmer jumps downward initially, which adds to the potential energy conversion. Use \( K.E._{initial} = \frac{1}{2}mv_0^2 \): \( v^2 = v_0^2 + 2gh \). Here, \( v_0 = 2.5 \mathrm{m/s} \) downward. Substitute to calculate \( v^2 = (2.50)^2 + 2 \times 9.8 \times 3.25 \) and solve for \( v \).

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Key Concepts

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

Potential Energy
Potential energy is stored energy that depends on an object's position or condition. In the context of our swimming exercise, it is the energy possessed by the swimmer due to their height above the water. Potential energy (PE) can be calculated using the formula:
  • \( PE = mgh \)
where:
  • \( m \) is the mass of the object, which is 72.0 kg for the swimmer,
  • \( g \) is the acceleration due to gravity, approximately 9.8 m/s虏 on Earth,
  • \( h \) is the height above the water, given as 3.25 m.
By substituting these values into the formula, you can determine the amount of stored energy before the swimmer dives. This energy is crucial since it gets transformed into kinetic energy when the swimmer jumps, showing the conservation of energy in action.
Kinetic Energy
Kinetic energy refers to the energy an object possesses due to its motion. When the swimmer in our exercise moves from the diving board to the water, potential energy transforms into kinetic energy. The kinetic energy (KE) of the swimmer at any point during the dive can be expressed as:
  • \( KE = \frac{1}{2}mv^2 \)
where:
  • \( m \) is the mass of the swimmer (72.0 kg),
  • \( v \) is the velocity or speed of the swimmer at that moment.
The process of conversion from potential to kinetic energy explains why the swimmer gains speed as they descend. The higher the initial height, the more potential energy, and thus more kinetic energy can be obtained, leading to a quicker descent as more energy is converted.
Mechanical Energy
Mechanical energy is the sum of an object's potential and kinetic energy. It represents the total energy available for the swimmer's motion and is governed by the principle of energy conservation. In our exercise, mechanical energy remains constant if we ignore air resistance. This means the sum of potential and kinetic energy at any point in the swimmer's dive equals the total mechanical energy initially present.Let's simplify things with an equation:
  • \( ME = PE + KE \)
At the top of the diving board, the swimmer's mechanical energy is all potential (since speed and thus kinetic energy are zero). As they fall, potential energy decreases while kinetic energy increases, keeping the mechanical energy the same.This principle helps calculate how fast the swimmer will move at different points, allowing us to solve each part of the exercise accurately by balancing potential and kinetic energy in the calculations.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) U.S. power use. The total consumption of electrical energy in the United States is about \(1.0 \times 10^{19}\) joules per year. (a) Express this rate in watts and kilowatts. (b) If the U.S. popu- lation is about 310 million people, what is the average rate of electrical energy consumption per person?

\(\bullet\) 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. Deter- mine whether he gives her a tender embrace or knocks her off her limb by calculating Tarzan's speed just before he reaches Jane. You can ignore air resistance and the mass of the vine.

\(\bullet\) \(\bullet\) \(\bullet\) A ball is thrown upward with an initial velocity of 15 \(\mathrm{m} / \mathrm{s}\) at an angle of \(60.0^{\circ}\) above the horizontal. Use energy conservation to find the ball's greatest height above the ground.

\(\bullet\) \(\bullet\) At the site of a wind farm in North Dakota, the average wind speed is \(9.3 \mathrm{m} / \mathrm{s},\) and the average density of air is 1.2 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) Calculate how much kinetic energy the wind contains, per cubic meter, at this location. (b) No wind turbine can capture all of the energy contained in the wind, the main reason being that capturing all the energy would require stop- ping the wind completely, meaning that air would stop flowing through the turbine. Suppose a particular turbine has blades with a radius of 41 \(\mathrm{m}\) and is able to capture 35\(\%\) of the avail- able wind energy. What would be the power output of this tur- bine, under average wind conditions?

\(\bullet\) \(\bullet\) A 20.0 kg rock slides on a rough horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average thermal power is produced as the rock stops?
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