91Ó°ÊÓ

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel)

Step by step solution

01

Understanding the Problem

We need to find the velocity of a ball when fired from a spring gun under different scenarios: (a) with no resistive force, (b) with a constant resistive force, and (c) find the position and maximum speed along the barrel given a resistive force.

02

Calculate Speed without Resistance (Case a)

Use energy conservation principles. The potential energy stored in the spring is given by\[ PE = \frac{1}{2} k x^2 \]where \( k = 400 \ \mathrm{N/m} \) and \( x = 0.06 \ \mathrm{m} \). The initial kinetic energy is zero, and all potential energy converts to kinetic energy.\[ PE = \frac{1}{2} \times 400 \times (0.06)^2 = 0.72 \ \mathrm{J} \]Set this equal to the kinetic energy, \( KE = \frac{1}{2} m v^2 \), and solve for \( v \):\[ 0.72 = \frac{1}{2} \times 0.03 \times v^2 \]\[ v^2 = \frac{0.72}{0.015} = 48 \]\[ v = \sqrt{48} \approx 6.93 \ \mathrm{m/s} \]

03

Calculate Speed with Resistance (Case b)

Account for work done against friction. The work done by the resistive force is\[ W = F \cdot d = 6 \ \mathrm{N} \times 0.06 \ \mathrm{m} = 0.36 \ \mathrm{J} \].The initial potential energy was 0.72 J, so the remaining energy is\[ KE = 0.72 \ \mathrm{J} - 0.36 \ \mathrm{J} = 0.36 \ \mathrm{J} \]Set this equal to the kinetic energy to find the velocity:\[ 0.36 = \frac{1}{2} \times 0.03 \times v^2 \]\[ v^2 = \frac{0.36}{0.015} = 24 \]\[ v = \sqrt{24} \approx 4.90 \ \mathrm{m/s} \]

04

Determine Position and Maximum Speed with Resistance (Case c)

The force of the spring at any point \( x \) is given by \( F = kx \). The resistive force is constant at 6 N. The net force is \( F_{net} = kx - 6 \). At maximum speed, acceleration is zero, and \( F_{net} = 0 \):\[ kx = 6 \]\[ x = \frac{6}{400} = 0.015 \ \mathrm{m} \].At \( x = 0.015 \) m, potential energy is\[ PE = \frac{1}{2} \times 400 \times (0.015)^2 = 0.045 \ \mathrm{J} \]The work done against the force from 0 to 0.015 m is\[ W = 6 \times 0.015 = 0.09 \ \mathrm{J} \].Initial potential energy was 0.72 J, so\[ KE = 0.72 - 0.09 - 0.045 = 0.585 \ \mathrm{J} \]Solve for velocity:\[ 0.585 = \frac{1}{2} \times 0.03 \times v^2 \]\[ v^2 = \frac{0.585}{0.015} = 39 \]\[ v = \sqrt{39} \approx 6.24 \ \mathrm{m/s} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Conservation of Energy

In physics, the conservation of energy principle states that energy cannot be created or destroyed, only transformed from one form to another. In the context of the spring gun problem, this principle plays a crucial role. When the spring is compressed, it stores potential energy. As the spring releases, this potential energy is converted into kinetic energy, which propels the ball out of the barrel. The initial potential energy stored in the spring is given by the formula:\[ PE = \frac{1}{2} k x^2 \]Here, \( k \) represents the spring constant, and \( x \) is the compression distance. This potential energy is then fully or partially converted into kinetic energy, depending on whether external forces like friction exist. Understanding how energy transitions from potential to kinetic is essential to solve problems involving moving objects and springs.

Spring Force

The spring force is one of the fundamental concepts in mechanics and plays a central role in this problem. It comes into play through Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement from the equilibrium position. This force is mathematically represented as:\[ F = -kx \]where \( F \) is the force exerted by the spring, \( k \) is the spring constant, and \( x \) is the displacement from equilibrium. The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. In this exercise, when the spring is compressed, it applies a force on the ball, which accelerates it down the barrel. This force is what initiates the conversion of potential energy into kinetic energy, propelling the ball outward.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. When the compressed spring is released, the stored potential energy in the spring transitions into the motion of the ball, thus converting into kinetic energy. The kinetic energy (KE) of the ball can be expressed with the formula: \[ KE = \frac{1}{2} mv^2 \]where \( m \) is the mass of the ball, and \( v \) is its velocity. The goal in many physics problems, like in this exercise, is to determine the velocity of the object once all the potential energy has been converted. In the case of no resistive forces, the maximum kinetic energy is reached at the moment the ball leaves the barrel, as all potential energy has transformed into kinetic energy. However, when resistive forces are present, this maximum kinetic energy is reduced, since part of the energy is spent overcoming these forces.

Resistive Force

Resistive forces are forces that oppose the motion of an object and can prevent the full conversion of potential to kinetic energy. In this exercise, a constant resistive force, like friction or air resistance, acts against the ball as it travels down the barrel. This resistive force is significant in real-world applications because it consumes some of the energy that would otherwise convert into motion, reducing the kinetic energy and thus the speed of the ball. The work done by resistive forces can be calculated using the formula: \[ W = F \cdot d \]where \( F \) is the magnitude of the force, and \( d \) is the displacement. In this scenario, the calculated work done by the resistive force is subtracted from the initial potential energy, leaving less energy available for conversion into kinetic energy, thereby altering the velocity of the ball.

Work and Energy

Work and energy are closely related concepts in physics, often joined under the work-energy principle. Work is defined as the process of energy transfer to an object via a force that causes the object to move. It is expressed as: \[ W = F \cdot d \cdot \cos(\theta) \]where \( F \) is the force applied, \( d \) is the displacement, and \( \theta \) is the angle between the force and the direction of displacement. In this exercise, the concept of work is used to calculate the energy change involved when the resistive force opposes the motion of the ball. This work done by the resistive force reduces the ball's kinetic energy, impacting its final speed as it exits the barrel. Understanding how work impacts energy transformation gives insight into solving problems where forces like friction reduce efficiency.