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A modest slap \(^{32}\) might consist of about \(1 \mathrm{~kg}\) of mass moving at \(2 \mathrm{~m} / \mathrm{s}\). How much kinetic energy is this?

Short Answer

Expert verified
Question: Calculate the kinetic energy of a slap with a mass of 1 kg and a velocity of 2 m/s. Answer: The kinetic energy of the slap is 2 Joules.

Step by step solution

01

Identify given values

First, we'll identify the given values for mass and velocity of the slap: - Mass (m) = 1 kg - Velocity (v) = 2 m/s
02

Apply the kinetic energy formula

Now, we will use the kinetic energy formula to calculate the kinetic energy of the slap: K = 0.5 * m * v^2
03

Substitute the given values

Next, we will substitute the given values for mass and velocity into the formula: K = 0.5 * (1 kg) * (2 m/s)^2
04

Solve for kinetic energy

Finally, we will solve the equation to find the kinetic energy: K = 0.5 * (1 kg) * (4 m^2/s^2) K = 0.5 * 4 kg * (m^2/s^2) K = 2 kg * (m^2/s^2) So, the kinetic energy of this modest slap is 2 Joules.

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Key Concepts

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

Physics Education
Understanding the principles of physics paves the way for a deeper appreciation of the laws that govern our universe. Particularly when it comes to energy, knowing how different forms of energy work is a foundational aspect in physics education.

For students, the topics of mechanical energy and the calculations for kinetic energy can seem challenging, but they are integral to understanding motion and the work-concept. Clarifying these concepts through the use of relatable examples, such as calculating the energy involved in a slap, helps make the content more engaging and easier to grasp.
Energy Concepts in Physics
The study of energy in physics is all about understanding its different forms and the laws that govern its transformation and conservation. Energy is present in various forms, such as kinetic, potential, thermal, and electrical energy.

The concept of mechanical energy is particularly important as it encompasses both kinetic and potential energy, which are vital for describing the movement and position of objects. Mechanical energy is often the main focus when studying the processes that occur in classical mechanics.
Mechanical Energy
Mechanical energy is the sum of the kinetic energy and potential energy present in a system. It reflects the ability of an object to do work due to its motion or position. For instance, when an object falls, its potential energy is converted into kinetic energy, reflecting a fundamental principle of mechanical energy: the conservation of energy.

In practical educational exercises, visualizing this transition - for example, from the potential energy of a raised object to the kinetic energy as it impacts the ground - aids in students' comprehension of these concepts.
Kinetic Energy Formula
The kinetic energy of an object can be quantified using the kinetic energy formula, which is defined as K = 0.5 * m * v2, where K stands for kinetic energy, m is the mass of the object, and v represents its velocity.

This formula is a critical tool for understanding how the velocity of an object is a significant factor in the amount of kinetic energy it has - a concept demonstrated in the slap example, where a 1 kg mass moving at 2 m/s has a kinetic energy of 2 Joules, showing that even modest speeds can result in tangible energy.

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Most popular questions from this chapter

Referring to Figure 12.7, examine performance at \(5 \mathrm{~m} / \mathrm{s}\) and at \(10 \mathrm{~m} / \mathrm{s}\), picking a representative power for each in the middle of the cluster of black points, and assigning a power value from the left-hand axis. What is the ratio of power values you read off the plot, and how does this compare to theoretical expectations for the ratio going like the cube of velocity?

Traveling down the road, you carefully watch a three-bladed wind turbine, determining that it takes two seconds to make a full revolution. Assuming it's operating near the peak of its efficiency curve \(^{41}\) according to Figure \(12.4\), how fast do you infer the wind speed to be if the blade length \(^{42}\) appears to be \(15 \mathrm{~m}\) long?

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Atmospheric pressure is about \(10^{5} \mathrm{~N} / \mathrm{m}^{2}\), meaning that a \(100,000 \mathrm{~N}\) weight of air-corresponding to a mass of \(10,000 \mathrm{~kg}\) -sits atop very square meter of the ground (at or near sea level). If the air density were constant at \(1.25 \mathrm{~kg} / \mathrm{m}^{3}\) - rather than decreasing with height as it actually does - how high would the atmosphere extend to result in this weight (mass)?

The largest wind turbines have rotor diameters \(^{40}\) around \(150 \mathrm{~m}\). Using a sensible efficiency of \(50 \%\), what power does such a jumbo turbine deliver at a maximum design wind speed of \(13 \mathrm{~m} / \mathrm{s} ?\)

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