/*! 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 59 A 4-lb collar can slide without ... [FREE SOLUTION] | 91Ó°ÊÓ

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A 4-lb collar can slide without friction along a horizontal rod and is released from rest at A. The undeformed lengths of springs BA and CA are 10 in. and 9 in., respectively, and the constant of each spring is 2800 lb/in. Determine the velocity of the collar when it has moved 1 in. to the right.

Short Answer

Expert verified
The collar's velocity is approximately 61.35 ft/s after moving 1 inch to the right.

Step by step solution

01

Understand the Problem

We're given a 4-lb collar sliding on a frictionless horizontal rod. The collar moves 1 inch to the right from point A. There are two springs attached, one at A (BA) with an undeformed length of 10 inches, and one at C (CA) with an undeformed length of 9 inches. Each spring has a spring constant of 2800 lb/in. We need to determine the collar's velocity after moving 1 inch right.
02

Calculate the Extension of the Springs

Initially, the spring BA is unstressed at 10 inches and the collar starts at point A. After moving 1 inch to the right, the length of spring BA becomes 11 inches, so the extension is 1 in. Similarly, spring CA is initially unstressed at 9 inches and becomes 8 inches after 1-inch movement (since point A is right of C), resulting in a compression of 1 inch.
03

Apply the Conservation of Energy Principle

Since there is no friction, the energy is conserved. Initially, the kinetic energy (KE) is zero because it starts from rest, and there is no potential energy in the springs as they are unstressed. When it moves 1 inch to the right, the potential energy (PE) stored in the springs can be calculated using \[ PE = \frac{1}{2} k (\Delta L)^2 \] for each spring, where \(k = 2800\) lb/in and \(\Delta L = 1\) inch.
04

Compute the Potential Energy in Springs

For spring BA and CA, the potential energy is calculated as \[ PE_B = \frac{1}{2} \times 2800 \times (1)^2 = 1400 \text{ lb-in} \] \[ PE_C = \frac{1}{2} \times 2800 \times (1)^2 = 1400 \text{ lb-in} \] The total potential energy is: \[ PE_{total} = PE_B + PE_C = 1400 + 1400 = 2800 \text{ lb-in} \].
05

Convert Energy Units

Since we operate in standard units, convert lb-in to ft-lb: \[ PE_{total} = 2800 \text{ lb-in} \times \frac{1 \text{ ft}}{12 \text{ in}} = 233.33 \text{ ft-lb} \].
06

Set up the Kinetic Energy Equation

The kinetic energy (KE) after the 1-inch move is given by \[ KE = \frac{1}{2}mv^2 \]. The potential energy converts into kinetic energy: \[ KE = PE_{total} \].
07

Solve for the Velocity

Substitute the mass in slugs (since weight = 4 lb, mass \(m = \frac{4}{32.2}\) slugs) and solve for \(v\) using: \[ \frac{1}{2} \times \frac{4}{32.2} \times v^2 = 233.33 \].Solving gives:\[ v^2 = \frac{233.33 \times 2 \times 32.2}{4} \]\[ v^2 = 3762.896 \]\[ v \approx 61.35 \text{ ft/s} \].

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is vital when studying moving objects like the collar in this problem. This law tells us that energy can change forms but cannot disappear or be created out of nothing.
In the realm of mechanical physics, we often deal with potential and kinetic energy. When a collar moves without friction, as described in our scenario, mechanical energy is conserved. At the start, the collar has zero kinetic energy because it is not moving. When the collar displaces, potential energy stored in the springs changes into kinetic energy.
The equation \[ \text{KE}_\text{initial} + \text{PE}_\text{initial} = \text{KE}_\text{final} + \text{PE}_\text{final} \] tells us that the total energy before and after the collar moves is the same due to energy conservation. Initially, there’s no potential energy because the springs are unstressed. Once the springs are stretched and compressed, they store energy. This energy converts into kinetic energy as the collar gains speed.
Spring Force
Springs hold potential energy through deformation. This energy follows Hooke's Law, which states that the force needed to compress or extend a spring is proportional to its deformation from the natural length.
This force is described with the formula \[ F = k \times \Delta L \] where \( F \) is the force from the spring, \( k \) is the spring constant, and \( \Delta L \) is the change in length.
The springs contribute to the system's potential energy when they are either compressed or extended. In the exercise, both springs are deformed by an inch when the collar shifts. The changes in spring lengths determine the energy stored, calculated using \[ \text{PE} = \frac{1}{2} k (\Delta L)^2 \].
For both springs, substituting \( k = 2800 \) lb/in and \( \Delta L = 1 \text{ inch} \) shows how springs directly relate to the collar's motion and energy change.
Velocity Calculation
Calculating velocity involves understanding how potential energy of the springs turns into kinetic energy of the moving collar. We can find velocity by equating the kinetic energy gained with the total potential energy from the springs: \[ \frac{1}{2} m v^2 = \text{PE}_{\text{total}} \].
This equation allows solving for velocity if mass and energy data are given.
In our scenario, after moving 1 inch, the total potential energy from decompressed springs is 233.33 ft-lb. Accounting for system parameters like the collar's weight of 4 lb (converted into mass using \( m = \frac{4}{32.2} \) slugs), you can solve for \( v \), the velocity.
The formula simplifies to \[ v^2 = \frac{233.33 \times 2 \times 32.2}{4} \], resulting in a calculated velocity of approximately 61.35 ft/s. This velocity indicates how changing potential energy from spring tension accelerates the collar along its path.

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Most popular questions from this chapter

A 2-kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when the collar is at C and its constant is 600 N/m. If the collar is released at A with no initial velocity, determine its velocity (a) as it passes through B, (b) as it reaches C.

The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur. Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200-g implant, determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest.

A \(1200-\mathrm{kg}\) trailer is hitched to a \(1400-\mathrm{kg}\) car. The car and trailer are traveling at \(72 \mathrm{km} / \mathrm{h}\) when the driver applies the brakes on both the car and the trailer. Knowing that the braking forces exerted on the car and the trailer are \(5000 \mathrm{N}\) and \(4000 \mathrm{N}\), respectively, determine \((a)\) the distance traveled by the car and trailer before they come to a stop, \((b)\) the horizontal component of the force exerted by the trailer hitch on the car.

Collar \(A\) has a mass of \(3 \mathrm{kg}\) and is attached to a spring of constant \(1200 \mathrm{N} / \mathrm{m}\) and of undeformed length equal to \(0.5 \mathrm{m} .\) The system is set in motion with \(r=0.3 \mathrm{m}, v_{\theta}=2 \mathrm{m} / \mathrm{s},\) and \(v_{r}=0 .\) Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when \(r=0.6 \mathrm{m} .\)

A spacecraft of mass \(m\) describes a circular orbit of radius \(r_{1}\) around the earth. \((a)\) Show that the additional energy \(\Delta E\) that must be imparted to the spacecraft to transfer it to a circular orbit of larger radius \(r_{2}\) is $$\Delta E=\frac{G M m\left(r_{2}-r_{1}\right)}{2 r_{1} r_{2}}$$ where \(M\) is the mass of the earth. ( \(b\) ) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path \(A B,\) the amounts of energy \(\Delta E_{A}\) and \(\Delta E_{B}\) which must be imparted at \(A\) and \(B\) are, respectively, proportional to \(r_{2}\) and \(r_{1}\) : $$\Delta E_{A}=\frac{r_{2}}{r_{1}+r_{2}} \Delta E \quad \Delta E_{B}=\frac{r_{1}}{r_{1}+r_{2}} \Delta E$$

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