A block of wood of mass 2 kg is placed over the muzzle of a rifle, and a bullet of mass 25 g is fired vertically upward. The bullet embeds in the wood, which rises to a macimum height of 3 m. Calculate the muzzle velocity of the bullet
A block of wood of mass 2 kg is placed over the muzzle of a rifle, and a bullet of mass 25 g is fired vertically upward. The bullet embeds in the wood, which rises to a macimum height of 3 m. Calculate the muzzle velocity of the bullet
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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A block of wood of mass 2 kg is placed over the muzzle of a rifle, and a bullet of mass 25 g is fired vertically upward. The bullet embeds in the wood, which rises to a macimum height of 3 m. Calculate the muzzle velocity of the bullet.
![**Example of a Hanging Object on a Spring**
In the provided illustration, we observe an example of a hanging object attached to a spring, commonly used in physics to demonstrate principles such as Hooke's Law and gravitational force.
**Description:**
1. **Object and Spring Arrangement:**
- There is an oblong-shaped object attached to a horizontal spring.
- The object appears to have a mass of 2 kg, as indicated in the diagram.
- The object is hanging freely under the influence of gravity.
- The spring is anchored to a fixed support at the top.
2. **Dimensions:**
- The distance between the top fixed support of the spring and the bottom of the mass is marked as 3 meters.
**Understanding the Setup:**
- This setup is often used to calculate the force exerted by the spring (spring force), which can be done using Hooke's Law:
\[
F = -kx
\]
Where:
\[
F \text{ is the force exerted by the spring}, \\
k \text{ is the spring constant}, \\
x \text{ is the extension or compression of the spring from its natural length}.
\]
- Gravitational force acting on the mass can be calculated using:
\[
F = mg
\]
Where:
\[
m \text{ is the mass}, \\
g \text{ is the acceleration due to gravity (approximately 9.8 m/s}^2\text{)}.
\]
In this case, the gravitational force acting on the 2 kg mass would be:
\[
F = 2 \text{ kg} \times 9.8 \text{ m/s}^2 = 19.6 \text{ N}
\]
- When the system is at equilibrium, the spring force equals the gravitational force, allowing the solving for variables such as the spring constant \( k \).
This diagram and its interpretation can help in understanding basic mechanical principles, particularly related to elastic forces, oscillations, and energy conservation in physics.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb870117d-bcad-4d3e-b30e-d1462d1fcc01%2F45cb1f1f-d548-42b8-a838-22482c8569dc%2Fz066qnl_processed.png&w=3840&q=75)
Transcribed Image Text:**Example of a Hanging Object on a Spring**
In the provided illustration, we observe an example of a hanging object attached to a spring, commonly used in physics to demonstrate principles such as Hooke's Law and gravitational force.
**Description:**
1. **Object and Spring Arrangement:**
- There is an oblong-shaped object attached to a horizontal spring.
- The object appears to have a mass of 2 kg, as indicated in the diagram.
- The object is hanging freely under the influence of gravity.
- The spring is anchored to a fixed support at the top.
2. **Dimensions:**
- The distance between the top fixed support of the spring and the bottom of the mass is marked as 3 meters.
**Understanding the Setup:**
- This setup is often used to calculate the force exerted by the spring (spring force), which can be done using Hooke's Law:
\[
F = -kx
\]
Where:
\[
F \text{ is the force exerted by the spring}, \\
k \text{ is the spring constant}, \\
x \text{ is the extension or compression of the spring from its natural length}.
\]
- Gravitational force acting on the mass can be calculated using:
\[
F = mg
\]
Where:
\[
m \text{ is the mass}, \\
g \text{ is the acceleration due to gravity (approximately 9.8 m/s}^2\text{)}.
\]
In this case, the gravitational force acting on the 2 kg mass would be:
\[
F = 2 \text{ kg} \times 9.8 \text{ m/s}^2 = 19.6 \text{ N}
\]
- When the system is at equilibrium, the spring force equals the gravitational force, allowing the solving for variables such as the spring constant \( k \).
This diagram and its interpretation can help in understanding basic mechanical principles, particularly related to elastic forces, oscillations, and energy conservation in physics.
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