/*! 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 309 The block of mass \(m\) is attac... [FREE SOLUTION] | 91Ó°ÊÓ

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The block of mass \(m\) is attached to the frame by the spring of stiffness \(k\) and moves horizontally with negligible friction within the frame. The frame and block are initially at rest with \(x=\) \(x_{0},\) the uncompressed length of the spring. If the frame is given a constant acceleration \(a_{0},\) determine the maximum velocity \(\dot{x}_{\max }=\left(v_{\mathrm{rel}}\right)_{\max }\) of the block relative to the frame.

Short Answer

Expert verified
\( \dot{x}_{max} = a_0 \cdot \sqrt{\frac{m}{k}} \)

Step by step solution

01

Understanding the System

Visualize the system consisting of a block of mass \(m\) connected to a frame by a spring of stiffness \(k\). Initially, both the block and frame are at rest with the spring being at its natural length \(x = x_0\). The frame starts moving with constant acceleration \(a_0\). Our goal is to find the maximum velocity of the block relative to the frame.
02

Determine the Maximum Relative Displacement

Due to the acceleration \(a_0\) of the frame, a pseudo-force \(m \cdot a_0\) will act on the block in the opposite direction. The block will move until the spring force \(k \cdot x\) equals this pseudo-force, i.e., \(k \cdot x = m \cdot a_0\). Thus, the maximum displacement of the block relative to the frame is \(x_{max} = \frac{m \cdot a_0}{k}\).
03

Use Energy Conservation to Find Maximum Velocity

The block achieves maximum velocity when it passes through equilibrium point. We use energy conservation considering only kinetic and spring potential energies:- Initial total energy (all potential due to pseudo-force): \(E_i = 0.5 \cdot k \cdot x_{max}^2 = 0.5 \cdot \frac{m^2 \cdot a_0^2}{k}\).- At maximum velocity relative to frame, the potential energy stored in the spring is zero, so total energy is all kinetic: \(E_f = 0.5 \cdot m \cdot \dot{x}_{max}^2\).Setting \(E_i = E_f\) gives \(0.5 \cdot m \cdot \dot{x}_{max}^2 = 0.5 \cdot \frac{m^2 \cdot a_0^2}{k}\).
04

Solve for Maximum Velocity

Solve the equation from the energy conservation step:\[ \dot{x}_{max} = \sqrt{\frac{m \cdot a_0^2}{k}} = a_0 \cdot \sqrt{\frac{m}{k}}. \]

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Key Concepts

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

Relative Motion
Relative motion helps us understand how an object moves in relation to another object or reference point. In this exercise, the problem involves a block and a frame. The block's motion is analyzed relative to the frame.
When something is at rest or moving together, relative motion might not seem significant. When the frame starts accelerating, the block appears to move relative to the frame. We can then measure how fast and how far the block travels from the perspective of the frame.
Understanding relative motion helps break down complex problems into simpler pieces. It allows us to focus on only one part of the system (in this case, the block), instead of both the block and the frame.
Spring-Mass System
A spring-mass system consists of a mass attached to a spring, and its behavior gives us insights into dynamic principles, such as oscillations and forces. In this scenario, the block of mass \(m\) is attached to a spring with stiffness \(k\).
Initially, the spring is at its natural length, meaning it is neither compressed nor stretched, so it exerts no force on the block. However, when the frame accelerates, a pseudo-force acts on the block, causing the spring to compress or stretch.
  • Spring stiffness \(k\) tells us how strong the spring is. A stiffer spring means more force is needed to compress it.
  • The spring force formula \(F = k \cdot x\) tells us the force exerted by the spring when it's displaced.
These principles help articulate how this system behaves under external influences such as the pseudo-force.
Energy Conservation
Energy conservation is a fundamental principle, stating energy cannot be created or destroyed, only transformed. In this context, the block of mass experiences both kinetic and potential energy changes.
Initially, all energy is stored as potential energy in the spring due to the pseudo-force acting on the block. This is calculated using the formula for potential energy in a spring: \(E_i = 0.5 \cdot k \cdot x_{max}^2\).
At maximum velocity, the energy is all kinetic because the potential energy is zero when the block moves through its equilibrium position, where the spring is momentarily at its natural length.
  • Kinetic energy is expressed as \(E_f = 0.5 \cdot m \cdot \dot{x}_{max}^2\).
Using the equality \(E_i = E_f\), we can solve for the maximum velocity, showing how energy transitions within the system are crucial to dynamics.
Pseudo-Force
A pseudo-force, or fictitious force, is an apparent force that arises when analyzing motion from a non-inertial frame of reference, like an accelerating frame.
In this exercise, when the frame accelerates, a pseudo-force \(m \cdot a_0\) acts on the block in the opposite direction of the frame’s motion. This occurs because the block wants to stay in its initial state due to inertia, but it "feels" a backward push relative to the accelerating frame.
This force influences the block as if a real force is applied, compressing or stretching the spring. Pseudo-forces can simplify dynamics problems since they allow us to apply Newton’s second law in accelerating reference frames, making complex, real-world scenarios more manageable.

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Most popular questions from this chapter

The figure represents the space shuttle \(S\), which is \((a)\) in a circular orbit about the earth and \((b)\) in an elliptical orbit where \(P\) is its perigee position. The exploded views on the right represent the cabin space with its \(x\) -axis oriented in the direction of the orbit. The astronauts conduct an experiment by applying a known force \(F\) in the \(x\) -direction to a small mass \(m .\) Explain why \(F=m \ddot{x}\) does or does not hold in each case, where \(x\) is measured within the spacecraft. Assume that the shuttle is between perigee and apogee in the elliptical orbit so that the orbital speed is changing with time. Note that the \(t-\) and \(x\) -axes are tangent to the path, and the \(\theta\) -axis is normal to the radial \(r\) -direction.

The 4 -kg ball and the attached light rod rotate in the vertical plane about the fixed axis at \(O .\) If the assembly is released from rest at \(\theta=0\) and moves under the action of the 60 -N force, which is maintained normal to the rod, determine the velocity \(v\) of the ball as \(\theta\) approaches \(90^{\circ} .\) Treat the ball as a particle.

The particle of mass \(m=1.2 \mathrm{kg}\) is attached to the end of the light rigid bar of length \(L=0.6 \mathrm{m} .\) The system is released from rest while in the horizontal position shown, at which the torsional spring is undeflected. The bar is then observed to rotate \(30^{\circ}\) before stopping momentarily. \((a)\) Determine the value of the torsional spring constant \(k_{T} .(b)\) For this value of \(k_{T},\) determine the speed \(v\) of the particle when \(\theta=15^{\circ}\).

Each tire on the 1350 -kg car can support a maximum friction force parallel to the road surface of 2500 N. This force limit is nearly constant over all possible rectilinear and curvilinear car motions and is attainable only if the car does not skid. Under this maximum braking, determine the total stopping distance \(s\) if the brakes are first applied at point \(A\) when the car speed is \(25 \mathrm{m} / \mathrm{s}\) and if the car follows the centerline of the road.

Two steel balls of the same diameter are connected by a rigid bar of negligible mass as shown and are dropped in the horizontal position from a height of \(150 \mathrm{mm}\) above the heavy steel and brass base plates. If the coefficient of restitution between the ball and the steel base is 0.6 and that between the other ball and the brass base is \(0.4,\) determine the angular velocity \(\omega\) of the bar immediately after impact. Assume that the two impacts are simultaneous.

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