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Some Typical Kinetic Energies. (a) In the Bohr model of the atom, the ground- state electron in hydrogen has an orbital speed of 2190 \(\mathrm{km} / \mathrm{s} .\) What is its kinetic energy? (Consult Appendix F.) (b) If you drop a 1.0-kg weight (about 2 lb) from a height of 1.0 \(\mathrm{m}\) , how many joules of kinetic energy will it have when it reaches the ground? (c) Is it reasonable that a \(30-\mathrm{kg}\) child could run fast enough to have 100 \(\mathrm{J}\) of kinetic energy?

Short Answer

Expert verified
(a) \(2.18 \times 10^{-18}\) J, (b) 9.8 J, (c) Yes, it's reasonable.

Step by step solution

01

Calculate the Kinetic Energy of the Electron

To find the kinetic energy (KE) of the electron, we use the formula for kinetic energy, which is \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the electron, \( 9.11 \times 10^{-31} \) kg, and \( v \) is the speed of the electron, 2190 km/s or \( 2.19 \times 10^6 \) m/s.Substituting these values into the formula gives:\[ KE = \frac{1}{2} \times 9.11 \times 10^{-31} \, \text{kg} \times (2.19 \times 10^6 \, \text{m/s})^2 \]Calculating this results in a kinetic energy of approximately \( 2.18 \times 10^{-18} \) joules.
02

Calculate the Kinetic Energy of the Falling Weight

Use the potential energy formula at height, which transforms into kinetic energy as it falls to the ground: \[ KE = mgh \]where \( m \) is the mass (1.0 kg), \( g \) is the acceleration due to gravity (9.8 m/s²), and \( h \) is the height (1.0 m).Substitute these into the formula:\[ KE = 1.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 \times 1.0 \, \text{m} \]This equals 9.8 joules of kinetic energy when it hits the ground.
03

Analyze the Kinetic Energy of the Running Child

For a child of mass 30 kg to reach a kinetic energy (KE) of 100 J, we use the kinetic energy formula: \[ KE = \frac{1}{2} mv^2 \]First, isolate \( v^2 \): \[ v^2 = \frac{2 \times KE}{m} \]Substitute the KE of 100 J and mass 30 kg:\[ v^2 = \frac{2 \times 100 \, \text{J}}{30 \, \text{kg}} = \frac{200}{30} = 6.67 \]To find the speed, take the square root of 6.67:\[ v \approx \sqrt{6.67} \approx 2.58 \, \text{m/s} \]This speed is reasonable for a child running.

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Key Concepts

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

Bohr model
The Bohr model is a foundational concept in atomic physics, which provides a simple way to describe the electron's motion around an atom's nucleus. In this model, electrons orbit the nucleus in circular paths or "shells." Each shell is associated with a quantized level of energy.
Bohr's revolutionary step was to propose that the electron's energy levels are quantized, meaning they can only have specific values. This quantization was a significant departure from classical physics.
In the Bohr model of the hydrogen atom, the ground-state electron possesses a specific energy determined by its orbit. The kinetic energy (KE) of this electron depends on its mass and velocity. Using the formula \[ KE = \frac{1}{2} mv^2 \], where:
  • \( m \) is the mass of the electron, approximately \( 9.11 \times 10^{-31} \) kg
  • \( v \) is the speed of the electron
This energy calculation shows how motion and energy levels are intricately connected in atomic structures.
potential energy
Potential energy is the stored energy of an object due to its position or state. In the context of physics, especially when analyzing motion and mechanics, potential energy often transforms into kinetic energy. This transformation is beautifully demonstrated when an object is dropped from a height, which is what happens in the classic 'falling weight' experiment.
Potential energy due to height for an object near the surface of the Earth can be calculated by the formula:\[ PE = mgh \] where:
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity, approximately \( 9.8 \text{ m/s}^2 \)
  • \( h \) is the height from which the object is dropped
As the object falls, this potential energy converts into kinetic energy (motion energy), illustrating the conservation of energy in a closed system, where total energy remains constant while changing its form.
mass and velocity relationships
Understanding the relationship between mass and velocity is fundamental when analyzing motion and energy, particularly in the study of kinetic energy. Kinetic energy is directly related to both mass and the square of velocity as per the formula:\[ KE = \frac{1}{2} mv^2 \]Here, we see how the mass \( m \) and velocity \( v \) of an object influence its energy.
  • More massive objects require more energy to move at the same velocity compared to less massive objects.
  • On the other hand, a small increase in velocity can lead to a significant increase in kinetic energy due to the squaring of velocity in the formula.
In practical scenarios, such as calculating the speed of a child running, understanding how much mass and velocity affect kinetic energy can provides insightful estimations of reasonable and achievable energy levels. This relationship exemplifies the broader notion of how changes in these variables impact energy transformations in physics.

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Most popular questions from this chapter

It takes a force of 53 \(\mathrm{kN}\) on the lead car of a 16 -car passenger train with mass \(9.1 \times 10^{5} \mathrm{kg}\) to pull it at a constant 45 \(\mathrm{m} / \mathrm{s}\) \((101 \mathrm{mi} / \mathrm{h})\) on level tracks. (a) What power must the locomotive provide to the lead car? (b) How much more power to the lead car than calculated in part (a) would be needed to give the train an acceleration of 1.5 \(\mathrm{m} / \mathrm{s}^{2}\) , at the instant that the train has a speed of 45 \(\mathrm{m} / \mathrm{s}\) on level tracks? (c) How much more power to the lead car than that calculated in part (a) would be needed to move the train up a 1.5\(\%\) grade (slope angle \(\alpha=\arctan 0.015 )\) at a constant 45 \(\mathrm{m} / \mathrm{s} ?\)

You throw a 20 -N rock vertically into the air from ground level. You observe that when it is 15.0 \(\mathrm{m}\) above the ground, it is trav- eling at 25.0 \(\mathrm{m} / \mathrm{s}\) upward. Use the work-energy theorem to find (a) the rock's speed just as it left the ground and (b) its maximum height.

A 20.0 -kg rock is sliding 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 power is produced by friction as the rock stops?

An ingenious bricklayer builds a device for shooting bricks up to the top of the wall where he is working. He places a brick on a vertical compressed spring with force constant \(k=450 \mathrm{N} / \mathrm{m}\) and negligible mass. When the spring is released, the brick is propelled upward. If the brick has mass 1.80 \(\mathrm{kg}\) and is to reach a maximum height of 3.6 \(\mathrm{m}\) above its initial position on the compressed spring, what distance must the bricklayer compress the spring initially? (The brick loses contact with the spring when the spring returns to its uncompressed length. Why?

Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?

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