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Chapter 10: Problem 26

A rifle fires a \(2.10 \times 10^{-2}\) kg pellet straight upward, because the pellet rests on a compressed spring that is released when the trigger is pulled. The spring has a negligible mass and is compressed by \(9.10 \times 10^{-2} \mathrm{~m}\) from its unstrained length. The pellet rises to a maximum height of \(6.10 \mathrm{~m}\) above its position on the compressed spring. Ignoring air resistance, determine the spring constant.

Short Answer

Expert verified
The spring constant is approximately 30.40 N/m.

Step by step solution

01

Analyze Given Data

We are provided with a pellet mass of \(m = 2.10 \times 10^{-2}\, \text{kg}\), maximum height \(h = 6.10\, \text{m}\), and initial compression of the spring \(x = 9.10 \times 10^{-2}\, \text{m}\). We need to find the spring constant \(k\).
02

Apply Energy Conservation Principle

Since the spring is released and the pellet reaches a maximum height, we can relate the potential energy from the spring to the gravitational potential energy at maximum height. This gives us the equation: \[ \frac{1}{2} k x^2 = mgh \] Where \(g = 9.8 \, \text{m/s}^2\) is the acceleration due to gravity.
03

Solve for Spring Constant

Rearrange the equation from Step 2 to solve for the spring constant \(k\): \[ k = \frac{2mgh}{x^2} \]
04

Substitute Known Values

Substitute \(m = 2.10 \times 10^{-2} \, \mathrm{kg}\), \(g = 9.8 \, \mathrm{m/s^2}\), \(h = 6.10 \, \mathrm{m}\), and \(x = 9.10 \times 10^{-2} \, \mathrm{m}\) into the formula for \(k\):\[ k = \frac{2 \times 2.10 \times 10^{-2} \times 9.8 \times 6.10}{(9.10 \times 10^{-2})^2} \]
05

Calculate the Spring Constant

Perform the calculations: \[ k = \frac{2 \times 2.10 \times 10^{-2} \times 9.8 \times 6.10}{8.281 \times 10^{-3}} = \frac{2.51676 \times 10^{-1}}{8.281 \times 10^{-3}} \approx 30.40 \, \text{N/m} \] The spring constant \(k\) is approximately \(30.40 \, \text{N/m}\).

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Key Concepts

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

Energy Conservation
In physics, energy conservation is a fundamental concept. It states that energy cannot be created or destroyed, only converted from one form to another. When the trigger is pulled, the stored energy in the compressed spring is converted to kinetic energy and then to gravitational potential energy as the pellet rises. This transformation helps us understand how different forms of energy interact in closed systems. It's like a magical exchange where energy seems to "travel" between forms, while the total amount remains constant!
Gravitational Potential Energy
Gravitational potential energy is energy an object possesses because of its position in a gravitational field. For our rifle pellet, the energy is highest at its peak height of 6.1 meters. This energy depends on three factors: mass, gravity, and height.
  • Mass ( m ) - The higher the mass, the more gravitational potential energy.
  • Gravity ( g ) - Constant at g = 9.8 \, \text{m/s}^2 on Earth's surface.
  • Height ( h ) - The higher the object, the more energy it has.
At maximum height, the kinetic energy is zero, and all the energy from the spring converts into gravitational potential energy: \[ mgh \].
Kinetic Energy
Kinetic energy is the energy of motion. When the spring is released, it pushes the pellet, converting potential energy stored in the spring into kinetic energy.
  • Kinetic energy depends on mass and velocity: \( \frac{1}{2} mv^2 \).
  • The moment the spring is fully uncompressed, it's when the pellet has maximum kinetic energy.
  • As the pellet rises, this kinetic energy is gradually converted into gravitational potential energy.
This relationship between motion and energy transformation is a core piece of understanding movement dynamics.
Spring Potential Energy
The spring potential energy is stored in the spring when compressed. It's a form of stored energy, ready to be transformed.
  • Spring energy is given by: \( \frac{1}{2} k x^2 \), where \( k \) is the spring constant and \( x \) is the compression distance.
  • Higher compression or a stiffer spring leads to more potential energy.
  • When released, this energy is converted to kinetic and then gravitational potential energy.
Understanding this energy gives insight into how springs can store and release energy efficiently.

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Most popular questions from this chapter

A point on the surface of a solid sphere (radius \(=R)\) is attached directly to a pivot on the ceiling. The sphere swings back and forth as a physical pendulum with a small amplitude. What is the length of a simple pendulum that has the same period as this physical pendulum? Give your answer in terms of \(R\).

A 70.0 -kg circus performer is fired from a cannon that is elevated at an angle of \(40.0^{\circ}\) above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by \(3.00 \mathrm{~m}\) from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as that of the net into which he is shot. He takes 2.14 s to travel the horizontal distance of \(26.8 \mathrm{~m}\) between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

Objects of equal mass are oscillating up and down in simple harmonic motion on two different vertical springs. The spring constant of spring 1 is \(174 \mathrm{~N} / \mathrm{m}\). The motion of the object on spring 1 has twice the amplitude as the motion of the object on spring \(2 .\) The magnitude of the maximum velocity is the same in each case. Find the spring constant of spring 2

Multiple-Concept Example 6 reviews the principles that play a role in this problem. A bungee jumper, whose mass is \(82 \mathrm{~kg}\), jumps from a tall platform. After reaching his lowest point, he continues to oscillate up and down, reaching the low point two more times in \(9.6 \mathrm{~s}\). Ignoring air resistance and assuming that the bungee cord is an ideal spring, determine its spring constant.

Interactive LearningWare 10.1 at reviews the concepts involved in this problem. A spring stretches by \(0.018 \mathrm{~m}\) when a \(2.8-\mathrm{kg}\) object is suspended from its end. How much mass should be attached to this spring so that its frequency of vibration is \(f=3.0\) \(\mathrm{Hz} ?\)
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