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Chapter 6: Problem 80

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.)

Short Answer

Expert verified
(a) 6.93 m/s; (b) 4.90 m/s; (c) Max speed 3 m/s at 1.5 cm from start.

Step by step solution

01

Understanding the Problem

We need to determine the speed of a ball as it leaves the barrel of a spring gun for two different conditions: without friction and with a constant resisting force. We also need to find where and what the maximum speed of the ball is along the barrel when there is a resisting force.
02

Converting Units

Convert the compression of the spring from centimeters to meters: 6.00 cm = 0.060 m.
03

Using Energy Conservation for Part (a)

For part (a), since there's no friction, we use the conservation of mechanical energy. The potential energy stored in the spring when compressed is given by \( PE = \frac{1}{2} k x^2 \), where \( x \) is the compression.Substitute \( k = 400 \) N/m and \( x = 0.060 \) m:\[ PE = \frac{1}{2} \times 400 \times (0.060)^2 = 0.72 \text{ J} \]This energy is converted into the kinetic energy \( KE = \frac{1}{2} m v^2 \) of the ball, where \( m = 0.0300 \) kg and \( v \) is the velocity.Set \( PE = KE \) and solve for \( v \):\[ 0.72 = \frac{1}{2} \times 0.0300 \times v^2 \]\[ v^2 = \frac{0.72}{0.015} = 48 \]\[ v = \sqrt{48} = 6.93 \text{ m/s} \]
04

Adding the Resisting Force for Part (b)

When a resisting force \( F = 6.00 \) N is present, some of the spring's potential energy is used to overcome this force. The work done by the resisting force is \( W = F \cdot d \), where \( d = 0.060 \) m.Calculate \( W \):\[ W = 6.00 \times 0.060 = 0.36 \text{ J} \]The remaining energy for the ball's kinetic energy is \( 0.72 - 0.36 = 0.36 \text{ J} \).Solve \( 0.36 = \frac{1}{2} \times 0.0300 \times v^2 \) for \( v \):\[ v^2 = \frac{0.36}{0.015} = 24 \]\[ v = \sqrt{24} = 4.90 \text{ m/s} \]
05

Finding Maximum Speed for Part (c)

The ball reaches maximum speed when the spring force equals the resisting force.The force exerted by the spring as a function of compression \( x \) is \( F_s = kx \).Set \( F_s = F_p \):\[ 400x = 6.00 \]\[ x = 0.015 \text{ m} \] (displacement from the equilibrium position)At this point, calculate the remaining spring energy to energy at \( x = 0.015 \text{ m} \):\[ PE_{spring} = \frac{1}{2} k x^2 = 0.5 \times 400 \times (0.045)^2 = 0.405 \text{ J} \]Use this in kinetic energy calculation as before to find maximum speed:\[ 0.405 - 6(0.045) = KE = 0.135 \text{ J} \]\[ v^2 = \frac{0.135}{0.015} = 9 \]\[ v = \,3 \text{ m/s} \] at \( x = 0.015 \text{ m} \).
06

Concluding Solution

Step through each scenario calculating the velocity using either energy conservation or energy with work principles, and pause to apply force considerations when needed for maximum speed.

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Key Concepts

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

Conservation of Energy
The concept of conservation of energy states that energy cannot be created or destroyed, only transformed from one form into another. In the context of our spring gun problem, mechanical energy is conserved because there is no friction acting on the ball during its motion.

Initially, the spring stores potential energy when compressed. This potential energy is calculated using the formula \[ PE = \frac{1}{2} k x^2 \], where \( k \) is the spring constant and \( x \) is the compression distance.

As the spring releases its stored energy, it is converted into the kinetic energy of the ball, calculated by \( KE = \frac{1}{2} m v^2 \), where \( m \) is the ball's mass and \( v \) is its velocity. Thus, all the spring's initial potential energy becomes the ball's kinetic energy when there is no friction or other forces acting on the ball.

This principle helps us calculate the ball's speed as it exits the barrel by setting the initial potential energy equal to the kinetic energy, leading to simplified energy transformation equations.
Work-Energy Principle
The work-energy principle is essential to understanding how forces, such as a constant resisting force, impact the ball's motion inside the spring gun's barrel.

According to this principle, the work done on an object is equal to the change in its kinetic energy. When there is a resisting force present, it does negative work on the ball by reducing the energy that can be converted into its kinetic energy.

We calculate the work done by this resisting force using the formula \[ W = F \cdot d \], where \( F \) is the force and \( d \) is the distance over which the force acts. In our problem, this reduces the ball's kinetic energy by the amount of work done, thereby reducing its exit speed.

This principle allows us to adjust the conservation of energy equation to \( PE - W = KE \), effectively accounting for both the energy converted into kinetic energy and the energy lost to the resisting force.
Frictionless Scenario
A frictionless scenario is an idealized condition that simplifies analyses by eliminating the effects of friction. In this problem, assuming a frictionless barrel allows us to solely focus on the forces acting due to the spring and any external forces like the resisting force.

Without friction, all the potential energy in the compressed spring is converted directly into the kinetic energy of the ball. This assumption makes it easier to calculate the speed of the ball as it leaves the barrel, using energy conservation without worrying about energy dissipated as heat due to friction.

Such scenarios are common in physics problems, helping to isolate specific elements and principles, such as the behavior of ideal springs and their energy transformations in mechanical systems.
Resisting Force
The presence of a resisting force is critical in determining the motion of the ball in the barrel of the spring gun. It represents any form of drag or resistance that slows the ball down, like air resistance or internal barrel friction.

In this scenario, a constant resisting force of 6.00 N opposes the spring force, slowing down the conversion of potential energy into kinetic energy. This force does work against the ball, decreasing its kinetic energy and consequently reducing its exit speed compared to a frictionless case.

To calculate the maximum speed in the presence of this resisting force, we note that the ball reaches its peak speed where the force from the spring matches the resisting force. At this point, any additional compression of the spring would only work against the resistance, without increasing the ball's speed, leading to a maximum who occurs before the ball exits the barrel.

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Most popular questions from this chapter

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Rescue. Your friend (mass 65.0 \(\mathrm{kg} )\) is standing on the ice in the middle of a frozen pond. There is very litle friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for 3.00 s and accelerate your friend from rest to a speed of 6.00 \(\mathrm{m} / \mathrm{s}\) whileyou remain at rest. What is the average power supplied by the force you applied?

BIO Chin-Ups, While doing a chin-up, a man lifts his body 0.40 \(\mathrm{m} .\) (a) How much work must the man do per kilogram of body mass? (b) The muscles involved in doing a chin-up can generate about 70 \(\mathrm{J}\) of work per kilogram of muscle mass. If the man can just barely do a \(0.40-\mathrm{m}\) chin-up, what percentage of his body's mass do these muscles constitute? (For comparison, the total percentage of muscle in a typical \(70-\mathrm{kg}\) man with 14\(\%\) body fat is about 43\(\%\) . (c) Repeat part (b) for the man's young son, who has arms half as long as his father's but whose muscles can also generate 70 \(\mathrm{J}\) of work per kilogram of muscle mass. (d) Adults and children have about the same percentage of muscle in their bodies. Explain why children can commonly do chin-ups more easily than their fathers.
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