The Stone Has A Mass Of 2 Kg. It Is Held 5 Meters Above The Ground And Then Dropped. Assume That $g = 10 \, \text{N/kg}$.(a) Calculate Its Kinetic Energy As The Stone Is About To Hit The Ground.(b) Calculate Its Speed As The Stone Is About To

Introduction

In this article, we will explore the physics of a stone being dropped from a height of 5 meters above the ground. We will calculate the kinetic energy of the stone as it is about to hit the ground and determine its speed at that moment. This problem involves the application of the concepts of potential energy, kinetic energy, and the acceleration due to gravity.

The Physics of the Problem

The stone has a mass of 2 kg and is held 5 meters above the ground. When it is dropped, it begins to accelerate downward due to the force of gravity. The acceleration due to gravity is given as $g = 10 \, \text{N/kg}$. We can use the equation for the potential energy of an object at a height $h$ above the ground:

$$U = mgh$$

where $m$ is the mass of the object, $g$ is the acceleration due to gravity, and $h$ is the height above the ground.

Calculating the Initial Potential Energy

The stone is initially at a height of 5 meters above the ground. We can calculate its initial potential energy using the equation above:

$$U_i = mgh_i = (2 \, \text{kg}) \times (10 \, \text{N/kg}) \times (5 \, \text{m}) = 100 \, \text{J}$$

The Conversion of Potential Energy to Kinetic Energy

As the stone falls, its potential energy is converted into kinetic energy. We can use the equation for the conservation of energy to relate the initial potential energy to the final kinetic energy:

$$U_i + K_i = U_f + K_f$$

Since the stone starts from rest, its initial kinetic energy is zero. Therefore, the equation simplifies to:

$$U_i = K_f$$

We can substitute the value of the initial potential energy into this equation to find the final kinetic energy:

$$K_f = U_i = 100 \, \text{J}$$

Calculating the Speed of the Stone

We can use the equation for the kinetic energy of an object to find its speed:

$$K = \frac{1}{2}mv^2$$

We can rearrange this equation to solve for the speed:

$$v = \sqrt{\frac{2K}{m}}$$

Substituting the values of the kinetic energy and the mass of the stone, we get:

$$v = \sqrt{\frac{2 \times 100 \, \text{J}}{2 \, \text{kg}}} = \sqrt{100 \, \text{m}^2/\text{s}^2} = 10 \, \text{m/s}$$

Conclusion

In this article, we calculated the kinetic energy of a stone as it is about to hit the ground and determined its speed at that moment. We used the concepts of potential energy, kinetic energy, and the acceleration due to gravity to solve the problem. The stone's kinetic energy was found to be 100 J, and its speed was determined to be 10 m/s.

References

• [1] Halliday, D., Resnick, R., & Walker, J. (2013). Fundamentals of Physics. John Wiley & Sons.
• [2] Serway, R. A., & Jewett, J. W. (2018). Physics for Scientists and Engineers. Cengage Learning.

Additional Resources

• [1] Khan Academy. (n.d.). Potential and Kinetic Energy. Retrieved from https://www.khanacademy.org/science/physics/energy-and-work/potential-and-kinetic-energy/v/potential-and-kinetic-energy
• [2] Physics Classroom. (n.d.). Potential Energy. Retrieved from https://www.physicsclassroom.com/class/energy/Lesson-1/Potential-Energy
  The Stone's Descent: Q&A ==========================

Q: What is the initial potential energy of the stone?

A: The initial potential energy of the stone is given by the equation $U_i = mgh_i$, where $m$ is the mass of the stone, $g$ is the acceleration due to gravity, and $h_i$ is the initial height above the ground. In this case, the initial potential energy is $U_i = (2 \, \text{kg}) \times (10 \, \text{N/kg}) \times (5 \, \text{m}) = 100 \, \text{J}$.

Q: What happens to the potential energy of the stone as it falls?

A: As the stone falls, its potential energy is converted into kinetic energy. This is a result of the conservation of energy, which states that the total energy of an isolated system remains constant over time.

Q: How do we calculate the kinetic energy of the stone as it is about to hit the ground?

A: We can use the equation for the conservation of energy to relate the initial potential energy to the final kinetic energy:

$$U_i + K_i = U_f + K_f$$

Since the stone starts from rest, its initial kinetic energy is zero. Therefore, the equation simplifies to:

$$U_i = K_f$$

We can substitute the value of the initial potential energy into this equation to find the final kinetic energy:

$$K_f = U_i = 100 \, \text{J}$$

Q: How do we calculate the speed of the stone as it is about to hit the ground?

A: We can use the equation for the kinetic energy of an object to find its speed:

$$K = \frac{1}{2}mv^2$$

We can rearrange this equation to solve for the speed:

$$v = \sqrt{\frac{2K}{m}}$$

Substituting the values of the kinetic energy and the mass of the stone, we get:

$$v = \sqrt{\frac{2 \times 100 \, \text{J}}{2 \, \text{kg}}} = \sqrt{100 \, \text{m}^2/\text{s}^2} = 10 \, \text{m/s}$$

Q: What is the relationship between the potential energy and the kinetic energy of the stone?

A: The potential energy and the kinetic energy of the stone are related by the conservation of energy. As the stone falls, its potential energy is converted into kinetic energy. This means that the total energy of the stone remains constant over time.

Q: What is the significance of the acceleration due to gravity in this problem?

A: The acceleration due to gravity is a fundamental constant that determines the rate at which the stone falls. In this problem, the acceleration due to gravity is given as $g = 10 \, \text{N/kg}$. This value is used to calculate the initial potential energy and the final kinetic energy of the stone.

Q: What are some real-world applications of the concepts of potential and kinetic energy?

A: The concepts of potential and kinetic energy have many real-world applications. For example, they are used to design and optimize systems such as roller coasters, water slides, and other amusement park attractions. They are also used in the design of buildings and bridges to ensure that they can withstand various types of loads and stresses.

Q: What are some common misconceptions about potential and kinetic energy?

A: One common misconception about potential and kinetic energy is that they are mutually exclusive. However, this is not the case. As an object falls, its potential energy is converted into kinetic energy. This means that the total energy of the object remains constant over time.

Q: How can I apply the concepts of potential and kinetic energy to my own life?

A: The concepts of potential and kinetic energy can be applied to many areas of life. For example, they can be used to optimize systems such as your daily routine, your finances, and your relationships. They can also be used to design and optimize your own personal projects and goals.