Problem 16
Question
A gun is fired and a \(50-\mathrm{g}\) bullet is accelerated to a muzzle speed of \(100 \mathrm{~m} / \mathrm{s}\). If the length of the gun barrel is \(0.90 \mathrm{~m}\), what is the magnitude of the accelerating force? (Assume the acceleration to be constant.)
Step-by-Step Solution
Verified Answer
The accelerating force is approximately 277.78 N.
1Step 1: Analyze the given information
The problem provides the mass of the bullet as 50 grams. We must convert this to kilograms for consistency in SI units, which gives us 0.05 kg. The muzzle speed (final velocity) is given as 100 m/s, and the distance in which this acceleration occurs is 0.90 m.
2Step 2: Apply the kinematic equation
To find the acceleration, we'll use the kinematic equation: \[ v^2 = u^2 + 2as \]where \( v = 100 \ m/s \) is the final velocity, \( u = 0 \ m/s \) is the initial velocity (since it starts from rest), and \( s = 0.90 \ m \) is the distance. Solving for \( a \), the equation becomes:\[ a = \frac{v^2}{2s} \]
3Step 3: Calculate the acceleration
Substitute the given values into the equation:\[ a = \frac{100^2}{2 \times 0.90} = \frac{10000}{1.8} \approx 5555.56 \ m/s^2 \]Thus, the acceleration is approximately 5555.56 m/s².
4Step 4: Use Newton's Second Law to find the force
Newton's Second Law states that \( F = ma \), where \( m \) is the mass in kilograms and \( a \) is the acceleration. Substituting the values:\[ F = 0.05 \times 5555.56 \approx 277.78 \ N \]
5Step 5: Conclude the magnitude of the force
The magnitude of the accelerating force on the bullet is approximately 277.78 N.
Key Concepts
Newton's Second LawAccelerationMuzzle Speed
Newton's Second Law
Newton's Second Law is a fundamental principle in classical mechanics. It explains how the motion of an object is changed by the forces acting on it. The law is typically written as \( F = ma \), where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced. This law tells us that the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass.
When a force is applied to a bullet in a gun barrel, such as in our exercise, Newton's Second Law helps us calculate the acceleration and the force needed to launch it at a specific speed. In this scenario, given the bullet's mass and the calculated acceleration, we can use \( F = ma \) to find the force exerted.
When a force is applied to a bullet in a gun barrel, such as in our exercise, Newton's Second Law helps us calculate the acceleration and the force needed to launch it at a specific speed. In this scenario, given the bullet's mass and the calculated acceleration, we can use \( F = ma \) to find the force exerted.
- The bullet's mass was given as 0.05 kg.
- An acceleration of about 5555.56 m/s² was calculated based on kinematic principles.
Acceleration
Acceleration is the rate at which an object's velocity changes over time. In the context of the problem, the bullet's acceleration is a measure of how quickly it goes from being stationary to moving at its muzzle speed.
The kinematic equation \( v^2 = u^2 + 2as \) helps determine this acceleration. Here, \( v \) is the final velocity (100 m/s), \( u \) is the initial velocity (0 m/s, since the bullet starts at rest), and \( s \) is the distance over which the acceleration occurs (0.9 m).
The kinematic equation \( v^2 = u^2 + 2as \) helps determine this acceleration. Here, \( v \) is the final velocity (100 m/s), \( u \) is the initial velocity (0 m/s, since the bullet starts at rest), and \( s \) is the distance over which the acceleration occurs (0.9 m).
- Rearrange this formula to solve for \( a \): \( a = \frac{v^2 - u^2}{2s} \).
- Substituting known values gives us \( a = \frac{100^2}{2 \times 0.90} \).
- This results in an acceleration of about 5555.56 m/s².
Muzzle Speed
Muzzle speed is the term used to describe the speed of a projectile at the moment it exits the muzzle of a gun. It's crucial for understanding the projectile's performance and behavior after leaving the barrel.
In the provided exercise, the muzzle speed is 100 m/s. This is the bullet's velocity as soon as it exits the gun's barrel, and it's achieved due to the constant acceleration it experiences over a distance of 0.9 meters.
In the provided exercise, the muzzle speed is 100 m/s. This is the bullet's velocity as soon as it exits the gun's barrel, and it's achieved due to the constant acceleration it experiences over a distance of 0.9 meters.
- A high muzzle speed implies the bullet leaves the barrel with a significant amount of kinetic energy.
- This speed is pivotal for determining the bullet's impact force and range.
Other exercises in this chapter
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