/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 32 A rifle fires a \(2.10 \times 10... [FREE SOLUTION] | 91Ó°ÊÓ

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A rifle fires a \(2.10 \times 10^{-2}-\mathrm{kg}\) 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 282.9 N/m.

Step by step solution

01

Identify energy conversion

When the spring is released, its potential energy is converted into the kinetic energy of the pellet, which is then converted into gravitational potential energy at the pellet's maximum height. We can use the conservation of energy principle here.
02

Determine initial potential energy in the spring

The potential energy stored in a compressed spring is given by \[ PE_{spring} = \frac{1}{2} k x^2 \]where \( k \) is the spring constant and \( x \) is the compression distance. Here, \( x = 9.10 \times 10^{-2} \ \text{m} \).
03

Determine gravitational potential energy at maximum height

The gravitational potential energy at the maximum height is \[ PE_{gravity} = mgh \]where \( m = 2.10 \times 10^{-2} \ \text{kg} \), \( g = 9.81 \ \text{m/s}^2 \), and \( h = 6.10 \ \text{m} \).
04

Apply conservation of energy

According to the conservation of energy, the initial potential energy in the spring equals the gravitational potential energy at the maximum height:\[ \frac{1}{2} k x^2 = mgh \]
05

Solve for the spring constant \(k\)

Substitute the given values into the equation:\[ \frac{1}{2} k (9.10 \times 10^{-2})^2 = (2.10 \times 10^{-2})(9.81)(6.10) \]Solve the equation for \( k \):\[ k = \frac{2 \times (2.10 \times 10^{-2} \times 9.81 \times 6.10)}{(9.10 \times 10^{-2})^2} \]Calculate \( k \) to find the spring constant value.

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Key Concepts

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

Conservation of Energy
Conservation of Energy is a principle stating that in isolated systems, energy can neither be created nor destroyed, only transformed from one form to another. Consider a spring-loaded rifle firing a pellet. Initially, energy is stored as potential energy in the compressed spring. As the spring releases, all of this potential energy transforms into kinetic energy, which propels the pellet upwards. Once the pellet breaks free from the spring, kinetic energy turns into gravitational potential energy as the pellet gains height. Here is what happens:
  • Initial: All energy is potential, stored in the compressed spring.
  • During motion: Potential energy is converted to kinetic energy of the moving pellet.
  • At the peak: Kinetic energy is fully transformed to gravitational potential energy.
In perfect conditions (like in our problem without air resistance), the sum of energies remains constant, ensuring the conservation of energy. This allows us to equate potential energy in the spring to the gravitational potential energy at the pellet's maximum height.
Potential Energy
Potential Energy is the energy held by an object because of its position or state. When we talk about a spring, potential energy is stored during its compression or stretching. For a compressed spring, the potential energy can be calculated using the formula:\[ PE_{spring} = \frac{1}{2} k x^2 \]where:
  • \(k\) is the spring constant, representing spring stiffness.
  • \(x\) is the displacement from its natural length in meters.
The formula tells us how much energy is stored based on how compressed the spring is. The stiffer the spring (higher \(k\)), the more energy it can store. It's this energy that launches the pellet upwards. Understanding the potential energy of the spring helps in knowing the power behind the pellet's launch and how it will convert into other energy types.
Kinetic Energy
Kinetic Energy is the energy of motion. When the spring releases its stored potential energy, this energy converts into kinetic energy, giving the pellet speed. The formula for kinetic energy is:\[ KE = \frac{1}{2} mv^2 \]where:
  • \(m\) is the mass of the pellet.
  • \(v\) is the velocity at which it moves.
As the pellet is shot upwards, kinetic energy is at its peak right after leaving the spring. As it continues to rise, kinetic energy decreases while gravitational potential energy increases. Even without explicit values of velocity, we understand that this transformation ensures that the pellet ascends until it reaches its maximum height. The conservation principle assures that as kinetic energy decreases, the energy isn't lost but merely converted into another form of energy in the system.

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Most popular questions from this chapter

Astronauts on a distant planet set up a simple pendulum of length 1.2 \(\mathrm{m}\) . The pendulum executes simple harmonic motion and makes 100 complete vibrations in 280 s. What is the magnitude of the acceleration due to gravity on this planet?

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) block is resting on a horizontal frictionless surface and is attached to a horizontal spring whose spring constant is 124 \(\mathrm{N} / \mathrm{m}\) . The block is shoved parallel to the spring axis and is given an initial speed of 8.00 \(\mathrm{m} / \mathrm{s}\) , while the spring is initially unstrained. What is the amplitude of the resulting simple harmonic motion?

A die is designed to punch holes with a radius of \(1.00 \times 10^{-2} \mathrm{m}\) in a metal sheet that is \(3.0 \times 10^{-3} \mathrm{m}\) thick, as the drawing illustrates. To punch through the sheet, the die must exert a shearing stress of \(3.5 \times 10^{8} \mathrm{Pa}\) . What force \(\overrightarrow{\mathrm{F}}\) must be applied to the die?

A vertical spring with a spring constant of 450 \(\mathrm{N} / \mathrm{m}\) is mounted on the floor. From directly above the spring, which is un- strained, a \(0.30-\mathrm{kg}\) block is dropped from rest. It collides with and sticks to the spring, which is compressed by 2.5 \(\mathrm{cm}\) in bringing the block to a momentary halt. Assuming air resistance is negligible, from what height (in \(\mathrm{cm} )\) above the compressed spring was the block dropped?

Multiple-Concept Example 11 explores the concepts that are important in this problem. Pendulum A is a physical pendulum made from a thin, rigid, and uniform rod whose length is \(d .\) One end of this rod is attached to the ceiling by a frictionless hinge, so the rod is free to swing back and forth. Pendulum B is a simple pendulum whose length is also d. Obtain the ratio \(T_{A} / T_{B}\) of their periods for small-angle oscillations.

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