/*! 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 31 A heavy-duty stapling gun uses a... [FREE SOLUTION] | 91Ó°ÊÓ

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A heavy-duty stapling gun uses a \(0.140-\mathrm{kg}\) metal rod that rams against the staple to eject it. The rod is attached to and pushed by a stiff spring called a "ram spring" \((k=32000 \mathrm{N} / \mathrm{m})\). The mass of this spring may be ignored. The ram spring is compressed by \(3.0 \times 10^{-2} \mathrm{m}\) from its unstrained length and then released from rest. Assuming that the ram spring is oriented vertically and is still compressed by \(0.8 \times 10^{-2} \mathrm{m}\) when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

Short Answer

Expert verified
The speed of the ram at the instant of contact is approximately 8.41 m/s.

Step by step solution

01

Identify Known Values and Unknown

Given:- Mass of the rod, \( m = 0.140 \text{ kg} \).- Spring constant, \( k = 32000 \text{ N/m} \).- Initial compression, \( x_i = 3.0 \times 10^{-2} \text{ m} \).- Final compression when hitting the staple, \( x_f = 0.8 \times 10^{-2} \text{ m} \).We need to find the speed \( v \) of the ram at the instant of contact with the staple.
02

Apply Conservation of Energy Principle

The total mechanical energy is conserved when the spring is released.Initially, all the energy is stored as potential energy in the spring:\[ E_i = \frac{1}{2} k x_i^2 \]Finally, the energy is a combination of kinetic energy of the rod and the potential energy of the spring when it is still compressed:\[ E_f = \frac{1}{2} m v^2 + \frac{1}{2} k x_f^2 \]
03

Set Initial Energy Equal to Final Energy

By the conservation of energy:\[ \frac{1}{2} k x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x_f^2 \]
04

Solve for Speed v

Rearrange the equation to solve for the speed \( v \):\[ \frac{1}{2} m v^2 = \frac{1}{2} k x_i^2 - \frac{1}{2} k x_f^2 \]Simplify:\[ \frac{1}{2} m v^2 = \frac{1}{2} k (x_i^2 - x_f^2) \]Solve for \( v \):\[ v^2 = \frac{k}{m} (x_i^2 - x_f^2) \]\[ v = \sqrt{\frac{k}{m} (x_i^2 - x_f^2)} \]
05

Substitute Values and Calculate

Substitute the given values into the equation to find \( v \):\[ v = \sqrt{\frac{32000}{0.140} ((3.0 \times 10^{-2})^2 - (0.8 \times 10^{-2})^2)} \]After calculating:\[ v \approx 8.41 \text{ m/s} \]

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Key Concepts

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

Spring Potential Energy
Spring potential energy is a key concept in understanding how energy is stored in elastic objects like springs. When you compress or stretch a spring from its natural length, it stores potential energy due to its deformation.

This potential energy can be calculated using the formula:
  • \[ E_s = \frac{1}{2} k x^2 \]
Here, \( E_s \) is the spring potential energy, \( k \) is the spring constant (which measures how stiff the spring is), and \( x \) is the amount of compression or extension from the spring's relaxed position.

In our stapling gun exercise, the crucial step is to understand how the spring's potential energy changes as it goes from its initial compression state \((x_i)\) to when it hits the staple with a final compression \((x_f)\). Initially, the spring stores maximum potential energy because it is compressed the most. As it moves towards ejecting the staple, some of this stored energy is converted into kinetic energy, which we'll explore next.
Kinetic Energy Calculation
Once the spring potential energy begins to convert, it turns into kinetic energy. Kinetic energy is the energy an object has due to its motion. It's an essential piece to solving dynamics and energy-related problems.

We calculate kinetic energy using the equation:
  • \[ E_k = \frac{1}{2} m v^2 \]
where \( E_k \) is the kinetic energy, \( m \) is the mass of the object in motion, and \( v \) is the velocity of the object.

In our problem, as the spring is released, the ram gains velocity, and its kinetic energy increases. This rise in kinetic energy corresponds with a decrease in the spring potential energy, emphasizing the shift of energy types without loss. The task is to find the speed \( v \) of the rod when it makes contact with the staple. By knowing how much energy the rod gains as kinetic energy, we can back-calculate to find its speed at that point of action.
Mechanical Energy Conservation
The principle of mechanical energy conservation is central in problems involving springs and kinetic motion, like our stapling gun exercise. This principle tells us that in the absence of non-conservative forces (like friction or air resistance), the total mechanical energy of a system remains constant.

Total mechanical energy is the sum of an object's kinetic and potential energy:
  • \[ E_{ ext{total}} = E_k + E_s \]
Initially, all energy in the system is stored as spring potential energy \(( E_i = \frac{1}{2} k x_i^2)\). As the system evolves, energy is transferred from potential to kinetic energy, so when the rod makes contact with the staple, the energy is split between reduced potential energy and kinetic energy. We see this relationship in the equation:
  • \[ \frac{1}{2} k x_i^2 = \frac{1}{2} m v^2 + \frac{1}{2} k x_f^2 \]
This equation helps us calculate the speed of the ram by equating initial and final energy states. By substituting known values, we solve for the desired quantity, helping us analyze how systems like spring-operated tools work effectively without an external energy input during the process.

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Most popular questions from this chapter

A 68.0 -kg bungee jumper is standing on a tall platform \(\left(h_{0}=\right.\) \(46.0 \mathrm{m}),\) as indicated in the figure. The bungee cord has a natural length of \(L_{0}=9.00 \mathrm{m}\) and, when stretched, behaves like an ideal spring with a spring constant of \(k=66.0 \mathrm{N} / \mathrm{m}\). The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. Concepts: (i) Can we use the conservation of mechanical energy to find his speed at any point along the descent? Explain your answer. (ii) What type of energy does he have when he is standing on the platform? (iii) What types of energy does he have at point A? (iv) What types of energy does he have at point \(\mathrm{B} ?\) Calculations: What is his speed when he is at the following heights above the water: (a) \(h_{\mathrm{A}}=37.0 \mathrm{m},\) and (b) \(h_{\mathrm{B}}=15.0 \mathrm{m} ?\)

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