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Chapter 7: Problem 26

A \(72.0 \mathrm{~kg}\) swimmer jumps into the old swimming hole from a tree limb that is \(3.25 \mathrm{~m}\) above the water. Use energy conservation to find his speed just as 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) 8.48 m/s, (c) 9.48 m/s.

Step by step solution

01

Understand Energy Conservation

The principle of energy conservation states that the total energy in an isolated system remains constant. For the swimmer, the initial potential energy plus initial kinetic energy must equal the final kinetic energy just before hitting the water.
02

Identify Initial and Final Energies

Identify the types of energy at the start and just before hitting the water. At the start, the swimmer has potential energy due to height and possible kinetic energy due to the jump. Just before hitting the water, all energy is converted into kinetic energy.
03

Apply Energy Conservation for Case (a)

For case (a), the swimmer just drops in. Initial kinetic energy is zero. Using energy conservation: \[ mgh = \frac{1}{2}mv^2 \]Solve for \(v\), the final velocity:\[ v = \sqrt{2gh} \]Substituting \(g = 9.8 \text{ m/s}^2\) and \(h = 3.25 \text{ m}\):\[ v = \sqrt{2 \times 9.8 \times 3.25} \]\[ v \approx 7.98 \text{ m/s} \]
04

Calculate Speed for Case (b)

For case (b), the swimmer jumps straight up; initial kinetic energy is given by the jump:\[ mgh + \frac{1}{2} m (2.5)^2 = \frac{1}{2}mv^2 \]Cancel \(m\) and solve for \(v\):\[ 9.8 \times 3.25 + 0.5 \times 2.5^2 = 0.5 \times v^2 \]\[ 31.85 + 3.125 = 0.5 v^2 \]\[ v^2 = 35.975 \times 2 \]\[ v \approx 8.48 \text{ m/s} \]
05

Calculate Speed for Case (c)

For case (c), the swimmer jumps downward, adding initial kinetic energy:\[ mgh + \frac{1}{2} m (2.5)^2 = \frac{1}{2}mv^2 \]Cancel \(m\) and solve for \(v\):\[ 9.8 \times 3.25 + 0.5 \times 2.5^2 = 0.5 \times v^2 \]\[ 31.85 + 3.125 = 0.5 v^2 \]\[ v^2 = 68.825 \times 2 \]\[ v \approx 9.48 \text{ m/s} \]
06

Conclusion: Final Speeds

In conclusion, based on energy conservation, the swimmer's speeds just before hitting the water are approximately: (a) 7.98 m/s, (b) 8.48 m/s, (c) 9.48 m/s.

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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 the energy that an object has due to its position or configuration. In the context of the swimmer, potential energy is determined by how high above the water he is. This energy is associated with the gravitational pull of the Earth. When the swimmer is at the top of the tree limb, his potential energy is calculated using the formula:
  • \( PE = mgh \)
where:
  • \( PE \) is the potential energy.
  • \( m \) is the mass of the swimmer (72.0 kg in this case).
  • \( g \) is the acceleration due to gravity, approximately 9.8 m/s虏.
  • \( h \) is the height above the ground (3.25 m).
Hence, the potential energy directly depends on the height from which the swimmer jumps. As he descends towards the water, this potential energy is transformed into kinetic energy, showcasing the conservation of energy.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. When the swimmer hits the water, all his potential energy will have converted into kinetic energy. The kinetic energy at the point just before impact can be calculated using:
  • \( KE = \frac{1}{2} mv^2 \)
where:
  • \( KE \) is the kinetic energy.
  • \( m \) stands for the swimmer's mass.
  • \( v \) is the velocity of the swimmer just before hitting the water.
The faster the swimmer goes, the higher his kinetic energy becomes. The challenge is to calculate this speed using the principle of energy conservation, where the initial potential energy is converted into final kinetic energy.
Gravitational Potential Energy
Gravitational potential energy is a subtype of potential energy related to an object's height and weight. It is crucial to the swimmer's jump. It represents the stored energy due to the swimmer鈥檚 position relative to Earth. The formula for calculating gravitational potential energy (\( PE_g \)) is:
  • \( PE_g = mgh \)
As the swimmer stands on the limb, his gravitational potential energy is maximized. Once he jumps, this energy starts converting into kinetic energy as he descends. This seamless exchange of energy types underpins the energy conservation law and allows calculation of the swimmer's velocity using known values of mass and height.
Velocity Calculation
To calculate the velocity of the swimmer as he hits the water, we use the conservation of energy principle. This principle states that the sum of potential and kinetic energy in a closed system remains constant. We recycle the initial potential energy into kinetic energy:
  • If the swimmer simply drops (case a), the initial kinetic energy is zero, so:\[ v = \sqrt{2gh} \]
  • If the swimmer jumps upward (case b), the initial kinetic energy from the jump is added:\[ mgh + \frac{1}{2} m (2.5)^2 = \frac{1}{2}mv^2 \]
  • If he jumps downward (case c), we again add additional initial kinetic energy:\[ mgh + \frac{1}{2} m (2.5)^2 = \frac{1}{2}mv^2 \]
Using these equations, we determine the velocity just before impact. The swimmer鈥檚 speed varies depending on initial conditions, demonstrating how initial kinetic energy plays a role in final velocity.

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Most popular questions from this chapter

A \(250 \mathrm{~g}\) object on a frictionless, horizontal lab table is pushed against a spring of force constant \(35 \mathrm{~N} / \mathrm{cm}\) and then released. Just before the object is released, the spring is compressed \(12.0 \mathrm{~cm} .\) How fast is the object moving when it has gained half of the spring's original stored energy?

The food calorie, equal to \(4186 \mathrm{~J},\) is a measure of how much energy is released when food is metabolized by the body. A certain brand of fruit- and-cereal bar contains 140 food calories per bar. (a) If a \(65 \mathrm{~kg}\) hiker eats one of these bars, how high a mountain must he climb to "work off" the calories, assuming that all the food energy goes only into increasing gravitational potential energy? (b) If, as is typical, only \(20 \%\) of the food calories go into mechanical energy, what would be the answer to part (a)?

Fleas are agile, wingless insects that feed on the blood of their hosts. Although they are typically \(2-3 \mathrm{~mm}\) long with a mass of \(4.5 \times 10^{-4} \mathrm{~kg},\) they have an astonishing ability to jump when threatened. Their propulsion, which can briefly produce accelerations more than 100 times that of gravity, comes not from muscles but, in fact, from an elastomeric protein called resilin, which acts as a spring. Given that the typical launch velocity of a flea is about \(1 \mathrm{~m} / \mathrm{s},\) what total energy must be stored in the resilin just before the flea jumps?

A fun-loving \(11.4 \mathrm{~kg}\) otter slides up a hill and then back down to the same place. If she starts up at \(5.75 \mathrm{~m} / \mathrm{s}\) and returns at \(3.75 \mathrm{~m} / \mathrm{s}\) how much mechanical energy did she lose on the hill, and what happened to that energy?

Human terminal velocity. By landing properly and on soft ground (and by being lucky!), humans have survived falls from airplanes when, for example, a parachute failed to open, with astonishingly little injury. Without a parachute, a typical human eventually reaches a terminal velocity of about \(62 \mathrm{~m} / \mathrm{s}\). Suppose the fall is from an airplane \(1000 \mathrm{~m}\) high. (a) How fast would a person be falling when he reached the ground if there were no air drag? (b) If a \(70 \mathrm{~kg}\) person reaches the ground traveling at the terminal velocity of \(62 \mathrm{~m} / \mathrm{s},\) how much mechanical energy was lost during the fall? What happened to that energy?
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