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Chapter 14: Problem 21

If you put \(127 \mathrm{~J}\) of work into throwing a \(0.123 \mathrm{~kg}\) rock vertically into the air without changing its internal energy, how high will it go?

Short Answer

Expert verified
The rock will reach a height of approximately 105 meters.

Step by step solution

01

Identify given values

The values given in the problem are: work done \(W = 127 \mathrm{~J}\), the mass of the stone \(m = 0.123 \mathrm{~kg}\), and the acceleration due to gravity \(g = 9.8 \mathrm{~m/s^2}\).
02

Understand the relationship between work and potential energy

At the highest point, the kinetic energy of the rock is zero, and all work done (\(W\)) is converted into potential energy (\(E_{\text{pot}}\)). Therefore, we have \(E_{\text{pot}} = W\).
03

Solve for the height

As we know that \(E_{\text{pot}} = mgh\), we can substitute \(E_{\text{pot}}\) with \(W\) and solve for \(h\). Rearranging the equation gives us \(h = \frac{W}{mg}\). Substituting the known values into the equation gives \(h = \frac{127}{0.123 \times 9.8}\)

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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 stored in an object due to its position relative to some reference point, usually the Earth. When you lift an object off the ground, it gains potential energy because it is now capable of doing work by falling back to the ground. The potential energy due to gravity is commonly expressed using the formula:
  • The formula is: \[ E_{\text{pot}} = mgh \] where \( m \) is the mass of the object, \( g \) is the acceleration due to gravity, and \( h \) is the height above the reference point.
  • The higher the object, the more potential energy it has.
  • Potential energy only depends on the vertical position (height) and not on the path taken to get there.
In the exercise, we use the concept of potential energy to determine how high a rock can be thrown by converting the work done on it into potential energy.
Work-Energy Principle
The work-energy principle is a concept in physics that relates the work done on an object to its energy. When work is done on an object, it can change the energy of that object. This could be in the form of kinetic energy, potential energy, or even other forms of energy.
  • Work is calculated using the formula: \[ W = Fd \] where \( F \) is the force applied, and \( d \) is the distance over which the force is applied.
  • The principle states that the total work done on an object is equal to the change in its mechanical energy.
  • In the given problem, the work done is converted entirely to potential energy once the rock reaches its peak height.
This principle helps us solve the exercise by equating the work done, stored as potential energy, to determine the height the rock travels.
Acceleration Due to Gravity
Acceleration due to gravity is a fundamental aspect of physics that describes the acceleration of an object due to the force of Earth's gravity. On Earth, this constant is usually denoted as \( g \) and has an approximate value of \( 9.8 \text{ m/s}^2 \).
  • It is the same for all objects, regardless of their mass, when air resistance is negligible.
  • This constant is vital in calculations for potential energy and projectile motion.
  • In the context of the exercise, \( g \) helps calculate how high the rock will go by determining the potential energy at its highest point.
Understanding this constant assists in understanding many concepts in physics involving motion and energy, making it a key figure in the problem-solving process.

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Most popular questions from this chapter

The convective heat transfer rate from a computer chip is \(3.00 \mathrm{~W}\). The surface temperature of the chip is \(28.0^{\circ} \mathrm{C}\) and the air temperature is \(18.0^{\circ} \mathrm{C}\). Determine the convective heat transfer coefficient for the chip.

A cooking pot on a stove has a surface area of \(0.01 \mathrm{~m}^{2}\). The surface temperature of the pot is \(70.0^{\circ} \mathrm{C}\) and the temperature of the surrounding air in the kitchen is \(20^{\circ} \mathrm{C}\). The convective heat transfer coefficient is \(12.0 \mathrm{~W} /\left(\mathrm{m}^{2} \cdot \mathrm{K}\right)\). Determine the convection heat transfer rate from the pot to the kitchen air.

What is the Reynolds number of water flowing at \(3.50 \mathrm{ft} / \mathrm{s}\) through a \(3.00\) inch diameter pipe? The density and viscosity of the water are \(62.4 \mathrm{lbm} / \mathrm{ft}^{3}\) and \(2.03 \times 10^{-5} \mathrm{lbf}-\mathrm{s} / \mathrm{ft}^{2}\) respectfully.

If the heat conduction rate through a \(3.00 \mathrm{~m}^{2}\) wall \(1.00 \mathrm{~cm}\) thick is \(37.9 \mathrm{~W}\) when the inside and outside temperatures are \(20.0\) and \(0.00^{\circ} \mathrm{C}\) respectively, determine the thermal conductivity of the wall.

The surface area of an average naked human adult is \(2.2 \mathrm{~m}^{2}\) and the average skin temperature is \(33^{\circ} \mathrm{C}\left(91^{\circ} \mathrm{F}\right)\). The emissivity of human skin is \(0.95\). Determine the radiation heat transfer rate from the human to the walls of a room if they are at \(22^{\circ} \mathrm{C}\left(7^{\circ} \mathrm{F}\right)\). This is why a naked person feels chilly at room temperature.
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