/*! 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 1 \( \mathrm{A}\) mass of \(5 \mat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 1

\( \mathrm{A}\) mass of \(5 \mathrm{lbm}\) is released when it is just in contact with a spring of stiffness 25 lbf/ft that is attached to the ground. What is the maximum spring compression? Compare this to the deflection if the mass was just resting on the compressed spring. What would be the maximum spring compression if the mass was released from a distance of 5 ft above the top of the spring?

Short Answer

Expert verified
The spring compression is zero when the mass is just in contact with the spring. The spring compression is \(M \cdot g / k\) when the mass is just resting on the compressed spring. The maximum spring compression when the mass is released from a distance of 5 ft above the spring is \(x = \sqrt{(2 \cdot M \cdot g \cdot h) / k}\)

Step by step solution

01

Understanding the given quantities

The given quantities are the mass \(M = 5 \, \mathrm{lbm}\), the spring stiffness \(k = 25 \, \mathrm{lbf/ft}\), and the gravitational constant \(g = 32.2 \, \mathrm{ft/s^2}\). The weight of the object is \(W = M \cdot g\). The mass in slugs (the English unit of mass) is \(M \, \mathrm{(in \, slugs)} = \frac{M \, (in \, lbm)}{g}.\)
02

Calculate the maximum compression when the mass is just in contact with the spring

The energy balance equation can be set up as follows: the initial potential energy \(PE = W \cdot h\) equals the sum of the kinetic energy \(KE = \frac{1}{2} \cdot m \cdot v^2\) and the elastic potential energy \(EPE = \frac{1}{2} \cdot k \cdot x^2\). Since the object is in static equilibrium when it is just in contact with the spring, its initial potential energy and kinetic energy are zero, so the equation simplifies to \(M \cdot g \cdot 0 = 0 + \frac{1}{2}\cdot k \cdot x^2\). Solving this equation for \(x\) leads to \(x = 0 \, \mathrm{ft}\), i.e., the spring compression is zero when the mass is just in contact with the spring.
03

Calculate the spring compression if the mass was just resting on the compressed spring

For this case, the mass is in static equilibrium, so the spring force \(k \cdot x\) equals the weight of the object \(W = M \cdot g\). Solving this equation for \(x\) leads to \(x = \frac{M \cdot g}{k}\).
04

Maximum spring compression if the mass was released from a distance of 5 ft above the top of the spring

Now the initial potential energy \(PE = M \cdot g \cdot h\) equals the sum of the kinetic energy \(KE = \frac{1}{2} \cdot m \cdot v^2 = 0\) and the elastic potential energy \(EPE = \frac{1}{2} k \cdot x^2\). The initial potential energy is now nonzero because the object is released from a height \(h = 5 \, \mathrm{ft}\) above the spring. The kinetic energy is zero because the object is released from rest. Setting up the equation gives \(M \cdot g \cdot h = 0 + \frac{1}{2}\cdot k \cdot x^2\). Solve this equation for \(x\) to find the maximum spring compression.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Inflate a toy balloon with air and release it. Watch as the balloon darts about the room. Explain what causes the phenomenon you see.

Viscous liquid from a circular tank, \(D=300 \mathrm{mm}\) in diameter, drains through a long circular tube of radius \(R=50 \mathrm{mm}\) The velocity profile at the tube discharge is \\[ u=u_{\max }\left[1-\left(\frac{r}{R}\right)^{2}\right] \\] Show that the average speed of flow in the drain tube is \(\bar{V}=\frac{1}{2} u_{\max } .\) Evaluate the rate of change of liquid level in the tank at the instant when \(u_{\max }=0.155 \mathrm{m} / \mathrm{s}\)

The narrow gap between two closely spaced circular plates initially is filled with incompressible liquid. At \(t=0\) the upper plate, initially \(h_{0}\) above the lower plate, begins to move downward toward the lower plate with constant speed, \(V_{0}\) causing the liquid to be squeezed from the narrow gap. Neglecting viscous effects and assuming uniform flow in the radial direction, develop an expression for the velocity field between the parallel plates, Hint: Apply conservation of mass to a control volume with the outer surface located at radius \(r\) Note that even though the speed of the upper plate is constant, the flow is unsteady. For \(V_{0}=0.01 \mathrm{m} / \mathrm{s}\) and \(h_{0}=2\) \(\mathrm{mm},\) find the velocity at the exit radius \(R=100 \mathrm{mm}\) at \(t=0\) and \(t=0.1\) s. Plot the exit velocity as a function of time, and explain the trend.

A small steel ball of radius \(r=1 \mathrm{mm}\) is placed on top of a horizontal pipe of outside radius \(R=50 \mathrm{mm}\) and begins to roll under the influence of gravity, Rolling resistance and air resistance are negligible. As the speed of the ball increases, it eventually leaves the surface of the pipe and becomes a projectile. Determine the speed and location at which the ball loses contact with the pipe.

Gases leaving the propulsion nozale of a rocket are modeled as flowing radially outward from a point upstream from the nozzle throat. Assume the speed of the exit flow, \(V_{e}\) has constant magnitude. Develop an expression for the axial thrust, \(T_{a}\) developed by flow leaving the nozzle exit plane. Compare your result to the one-dimensional approximation, \(T=m v_{e^{-}}\) Evaluate the percent error for \(\alpha=15^{\circ} .\) Plot the percent error versus a for \(0
See all solutions

Recommended explanations on Physics Textbooks

Electromagnetism

Read Explanation

Scientific Method Physics

Read Explanation

Medical Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Energy Physics

Read Explanation

Atoms and Radioactivity

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.