A 10 Kg Ball Is Released From The Top Of A Hill. How Fast Is The Ball Going When It Reaches The Base Of The Hill?Approximate $g$ As $10 , \text{m/s}^2$ And Round The Answer To The Nearest Tenth. □ M/s \square \, \text{m/s} □ M/s

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Introduction

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In this article, we will explore the concept of potential and kinetic energy, and how it relates to the motion of an object. We will use the example of a 10 kg ball released from the top of a hill to calculate its final velocity when it reaches the base of the hill.

Understanding the Problem

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The problem states that a 10 kg ball is released from the top of a hill. We are asked to find the final velocity of the ball when it reaches the base of the hill. To solve this problem, we need to use the concept of potential and kinetic energy.

Potential and Kinetic Energy

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Potential energy is the energy an object has due to its position or configuration. In this case, the ball has potential energy due to its height above the ground. Kinetic energy, on the other hand, is the energy an object has due to its motion.

The formula for potential energy is:

$$U = mgh$$

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

The formula for kinetic energy is:

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

where $K$ is the kinetic energy, $m$ is the mass of the object, and $v$ is the velocity of the object.

Conservation of Energy

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The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another. In this case, the potential energy of the ball at the top of the hill is converted to kinetic energy as it rolls down the hill.

We can use the law of conservation of energy to set up an equation:

$$U = K$$

Substituting the formulas for potential and kinetic energy, we get:

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

Solving for Velocity

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We can solve for velocity by rearranging the equation:

$$v^2 = 2gh$$

Taking the square root of both sides, we get:

$$v = \sqrt{2gh}$$

Plugging in Values

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We are given that the mass of the ball is 10 kg, and we will approximate the acceleration due to gravity as 10 m/s^2. We need to find the height of the hill, which is not given. However, we can assume a typical height for a hill, such as 10 meters.

Plugging in the values, we get:

$$v = \sqrt{2 \times 10 \times 10}$$

$$v = \sqrt{200}$$

$$v = 14.14 \, \text{m/s}$$

Rounding to the Nearest Tenth

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We are asked to round the answer to the nearest tenth. Therefore, the final velocity of the ball when it reaches the base of the hill is approximately 14.1 m/s.

Conclusion

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In this article, we used the concept of potential and kinetic energy to calculate the final velocity of a 10 kg ball released from the top of a hill. We assumed a typical height for the hill and approximated the acceleration due to gravity. The final velocity of the ball when it reaches the base of the hill is approximately 14.1 m/s.

Frequently Asked Questions

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Q: What is the formula for potential energy?

A: The formula for potential energy is:

$$U = mgh$$

Q: What is the formula for kinetic energy?

A: The formula for kinetic energy is:

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

Q: What is the law of conservation of energy?

A: The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another.

Q: How do we solve for velocity?

A: We can solve for velocity by rearranging the equation:

$$v^2 = 2gh$$

Taking the square root of both sides, we get:

$$v = \sqrt{2gh}$$

Q: What is the final velocity of the ball when it reaches the base of the hill?

A: The final velocity of the ball when it reaches the base of the hill is approximately 14.1 m/s.

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Introduction

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In our previous article, we explored the concept of potential and kinetic energy, and how it relates to the motion of an object. We used the example of a 10 kg ball released from the top of a hill to calculate its final velocity when it reaches the base of the hill. In this article, we will answer some frequently asked questions related to this topic.

Q&A

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Q: What is the formula for potential energy?

A: The formula for potential energy is:

$$U = mgh$$

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

Q: What is the formula for kinetic energy?

A: The formula for kinetic energy is:

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

where $K$ is the kinetic energy, $m$ is the mass of the object, and $v$ is the velocity of the object.

Q: What is the law of conservation of energy?

A: The law of conservation of energy states that energy cannot be created or destroyed, only converted from one form to another.

Q: How do we solve for velocity?

A: We can solve for velocity by rearranging the equation:

$$v^2 = 2gh$$

Taking the square root of both sides, we get:

$$v = \sqrt{2gh}$$

Q: What is the final velocity of the ball when it reaches the base of the hill?

A: The final velocity of the ball when it reaches the base of the hill is approximately 14.1 m/s.

Q: What is the effect of gravity on the ball's motion?

A: Gravity is the force that pulls the ball towards the ground, causing it to accelerate downward. As the ball rolls down the hill, its potential energy is converted to kinetic energy, and its velocity increases due to the acceleration caused by gravity.

Q: Can we assume a constant acceleration due to gravity?

A: Yes, we can assume a constant acceleration due to gravity, which is approximately 10 m/s^2 on Earth.

Q: How does the mass of the ball affect its motion?

A: The mass of the ball does not affect its motion in this scenario, as we are assuming a constant acceleration due to gravity. However, if the ball were to experience a non-uniform acceleration, its mass would play a role in determining its motion.

Q: Can we use this formula to calculate the velocity of an object on a different planet?

A: Yes, we can use this formula to calculate the velocity of an object on a different planet, as long as we know the acceleration due to gravity on that planet.

Conclusion

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In this article, we answered some frequently asked questions related to the motion of a 10 kg ball released from the top of a hill. We covered topics such as potential and kinetic energy, the law of conservation of energy, and the effect of gravity on the ball's motion.

Frequently Asked Questions (FAQs)

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Q: What is the difference between potential and kinetic energy?

A: Potential energy is the energy an object has due to its position or configuration, while kinetic energy is the energy an object has due to its motion.

Q: Can we use this formula to calculate the velocity of an object on a different planet?

A: Yes, we can use this formula to calculate the velocity of an object on a different planet, as long as we know the acceleration due to gravity on that planet.

Q: How does the mass of the ball affect its motion?

A: The mass of the ball does not affect its motion in this scenario, as we are assuming a constant acceleration due to gravity.

Q: Can we assume a constant acceleration due to gravity?

A: Yes, we can assume a constant acceleration due to gravity, which is approximately 10 m/s^2 on Earth.

Q: What is the effect of gravity on the ball's motion?

A: Gravity is the force that pulls the ball towards the ground, causing it to accelerate downward.

Additional Resources

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• Wikipedia: Potential Energy
• Wikipedia: Kinetic Energy
• Wikipedia: Law of Conservation of Energy
• Wikipedia: Acceleration Due to Gravity