/*! 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 26 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 304 N/m.

Step by step solution

01

Identify and List Known Quantities

First, note the given quantities: - Mass of the pellet, \( m = 2.10 \times 10^{-2} \text{ kg} \)- Compression distance of the spring, \( x = 9.10 \times 10^{-2} \text{ m} \)- Maximum height reached by the pellet, \( h = 6.10 \text{ m} \).
02

Apply Conservation of Energy Principle

Use the conservation of mechanical energy where the potential energy stored in the spring is converted to gravitational potential energy at maximum height. Express the equation:\[ \frac{1}{2} k x^2 = mgh \]where \( k \) is the spring constant and \( g = 9.81 \text{ m/s}^2 \) is the acceleration due to gravity.
03

Solve for the Spring Constant

Rearrange the equation from the previous step to solve for the spring constant \( k \): \[ k = \frac{2mgh}{x^2} \]Substitute the values to find \( k \): \[ k = \frac{2 \cdot (2.10 \times 10^{-2}) \cdot 9.81 \cdot 6.10}{(9.10 \times 10^{-2})^2} \]
04

Perform the Calculation

Calculate \( k \) using the provided values:\[ k = \frac{2 \cdot 2.10 \times 10^{-2} \cdot 9.81 \cdot 6.10}{(9.10 \times 10^{-2})^2} = \frac{2.52}{0.008281} \approx 304 \text{ N/m} \].

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Key Concepts

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

Spring Constant
The spring constant is a crucial part of understanding how springs behave in physics. It tells us how stiff or flexible a spring is. The spring constant, often denoted as \( k \), is measured in newtons per meter (N/m). It reflects the force needed to compress or extend the spring by one meter.

In the context of the exercise, we are dealing with a compressed spring that launches a pellet upward. The compression of the spring stores potential energy, which is then converted into kinetic energy as the spring is released, propelling the pellet into the air.

To find the spring constant, we use the energy conservation principle that relates the stored potential energy in the spring to the gravitational potential energy at the highest point the pellet reaches. The formula used is:
  • Stored Energy in the Spring: \( \frac{1}{2} k x^2 \)
  • Gravitational Potential Energy: \( mgh \)
By finding \( k \), we reveal how efficiently the spring can perform its intended function: converting its stored energy into motion.
Gravitational Potential Energy
Gravitational potential energy is the energy that an object possesses because of its position in a gravitational field. It's like lifting a ball; higher positions mean more energy stored. The formula for gravitational potential energy is \( mgh \), where \( m \) is mass, \( g \) is the acceleration due to gravity (9.81 m/s² on Earth), and \( h \) is the height.

In our exercise, the pellet reaches a height of 6.10 meters after being launched from the spring. This height represents the maximum potential energy the pellet achieves thanks to the spring's stored power being transferred to it.

As the pellet gains height, it trades kinetic energy for gravitational potential energy until it reaches its peak. https://ru.bookmate.com/books/TkQ2BVHJ
  • This energy can be calculated using the known mass of the pellet, gravity, and the achieved height.
Understanding this exchange of energy forms is crucial for knowing how energy conserves and transforms in physical systems.
Mechanical Energy
Mechanical energy is the sum of potential and kinetic energy in a system, one of the basic principles in physics. It includes various forms of energy in motion and stored energy ready to be used. The conservation of mechanical energy is key in systems without external forces like friction or air resistance.

For the exercise, mechanical energy conservation means that all the energy the spring stored must be transferred to the pellet without loss. This spring-generated energy propels the pellet to its peak. We see this as:
  • The initial energy in the spring, given by \( \frac{1}{2} k x^2 \),
  • becomes the potential energy at the pellet's highest point, \( mgh \).
In ideal scenarios, characterized by perfect conservation, the total mechanical energy remains the same before and after events like shooting the pellet. It provides a way to calculate unknowns like the spring constant or the potential position energy just by considering how energy shifts and conserves in the system.

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Most popular questions from this chapter

Between each pair of vertebrae in the spinal column is a cylindrical disc of cartilage. Typically, this disc has a radius of about \(3.0 \times 10^{-2} \mathrm{~m}\) and a thickness of about \(7.0 \times 10^{-3} \mathrm{~m}\). The shear modulus of cartilage is \(1.2 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}\). Suppose a shearing force of magnitude \(11 \mathrm{~N}\) is applied parallel to the top surface of the disc while the bottom surface remains fixed in place. How far does the top surface move relative to the bottom surface?

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 \(\overrightarrow{\mathbf{F}}\) must be applied to the die?

A \(0.70-\mathrm{kg}\) block is hung from and stretches a spring that is attached to the ceiling. A second block is attached to the first one, and the amount that the spring stretches from its unstrained length triples. What is the mass of the second block?

A \(15.0\) -kg block rests on a horizontal table and is attached to one end of a massless, horizontal spring. By pulling horizontally on the other end of the spring, someone causes the block to accelerate uniformly and reach a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) in \(0.500 \mathrm{~s}\). In the process, the spring is stretched by \(0.200 \mathrm{~m}\). The block is then pulled at a constant speed of \(5.00 \mathrm{~m} / \mathrm{s}\), during which time the spring is stretched by only \(0.0500 \mathrm{~m}\). Find (a) the spring constant of the spring and (b) the coefficient of kinetic friction between the block and the table.

Interactive LearningWare \(10.2\) at explores the approach taken in problems such as this one. A spring is mounted vertically on the floor. The mass of the spring is negligible. A certain object is placed on the spring to compress it. When the object is pushed further down by just a bit and then released, one up/down oscillation cycle occurs in \(0.250 \mathrm{~s}\). However, when the object is pushed down by \(5.00 \times 10^{-2} \mathrm{~m}\) to point \(P\) and then released, the object flies entirely off the spring. To what height above point \(P\) does the object rise in the absence of air resistance?

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