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Chapter 5: Problem 8

Follow the evolution of energy exchanges for a wad of clay that you throw high into the air. Describe what is happening as the clay moves upward, as it reaches its apex, as it falls back down, and finally hits the ground with a thud. Where does the initial energy you put into the clay end up?

Short Answer

Expert verified
Answer: The three types of energy involved in the trajectory of a wad of clay thrown into the air are kinetic energy (KE), potential energy (PE), and internal energy (IE). The initial energy put into the clay, which starts as kinetic energy, ends up as internal energy within the clay due to the deformation and slight heating when it hits the ground.

Step by step solution

01

Identify the energy types involved in the problem

To follow the energy exchanges in this exercise, we'll need to consider three types of energy: - Kinetic energy (KE): The energy due to the clay's motion - Potential energy (PE): The energy due to the clay's position in the gravitational field - Internal energy (IE): The energy stored within the clay due to molecular bonds and interactions (associated with the clay deformation when it hits the ground)
02

As the clay moves upward

When the clay is thrown upward, its kinetic energy (KE) is converted into potential energy (PE) due to gravity. The kinetic energy decreases as the clay moves higher, and at the same time, the potential energy increases. The total mechanical energy (TME) of the system (KE + PE) remains the same, as long as we disregard any air resistance.
03

As the clay reaches its apex

At the apex of the clay's trajectory, its velocity momentarily becomes zero, meaning its kinetic energy (KE) is also zero. At this point, all of the initial kinetic energy has been converted into gravitational potential energy (PE). The total mechanical energy (TME) of the system remains constant (assuming no air resistance).
04

As the clay falls back down

As the clay falls back to the ground, its potential energy (PE) is converted back into kinetic energy (KE). The potential energy decreases while the kinetic energy increases, and the total mechanical energy (TME) of the system remains constant (again, assuming no air resistance).
05

As the clay hits the ground

When the clay impacts the ground, its kinetic energy is quickly transferred to internal energy (IE) within the clay as it deforms and heats up slightly. The kinetic energy goes to zero as the clay comes to a stop, and the potential energy goes to zero as well, since it is on the ground.
06

Determine where the initial energy ends up

The initial energy that was put into the clay (as kinetic energy) ends up as internal energy within the clay after it hits the ground. This is due to the deformation and slight heating of the clay upon impact.

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Most popular questions from this chapter

A \(50 \mathrm{~kg}\) crate might require \(200 \mathrm{~N}\) to slide across a concrete floor. If we must slide it \(10 \mathrm{~m}\) along the floor and then lift it \(2 \mathrm{~m}\) into a truck, how much energy goes into each action, and what fraction of the total energy expenditure is each?

The numbers in Table \(5.1\) are reasonable but should not be thought of as right. \(^{40}\) You can make your own table by using \(m g h\) for lifting and \(\frac{1}{2} m v^{2}\) for kinetic energy. For this exercise, pick three familiar activities or situations that allow you to estimate an energy scale in Joules and compute/estimate the results.

Heating a typical house might require something like \(200 \mathrm{~W}\) of power for every degree Celsius difference between inside and outside temperatures. If the inside temperature is kept at \(20^{\circ} \mathrm{C}\) and the outside temperature holds steady all day and night at \(0^{\circ} \mathrm{C}_{t}\) how much power is required to maintain the temperature?

Gather up or compute conversion factors from the chapter to start your own conversion table (empty version below). Express kWh, cal, kcal, Btu, and Therms in terms of Joules.

If Problem 21 had resulted in \(5,000 \mathrm{~W},{ }^{46}\) how many kilowatt- hours (kWh) are expended in a 24 -hour period? At an electricity cost of around \(\$ 0.15\) per \(\mathrm{kWh},{ }^{47}\) about how much will it cost, per day, to maintain heat?
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