**Problem Statement:**A mass weighing 24 lbs stretches a spring 2 inches. The mass is in a medium that exerts a viscous resistance of 28 lbs when the mass has a velocity of 2 ft/sec.Suppose the object is displaced an additional 6 inches and released.Find an equation for the object's displacement, \( u(t) \), in feet after \( t \) seconds.\[ u(t) = \] (with a blank placeholder for the equation)---### Analysis and Problem-Solving Steps:1. **Determine Constants:** - **Spring Constant (k):** The weight of the mass, \( W = 24 \) lbs, stretches the spring by 2 inches \( = \frac{2}{12} \) ft \( = \frac{1}{6} \) ft. Using Hooke's Law: \( W = kx \), where \( x \) is the displacement, \[ 24 = k \left( \frac{1}{6} \right) \] Solving for \( k \): \[ k = 24 \times 6 = 144 \text{ lbs/ft} \] - **Damping Coefficient (c):** The viscous resistance is \( 28 \) lbs at \( 2 \) ft/sec, \[ c \cdot 2 = 28 \] Solving for \( c \): \[ c = \frac{28}{2} = 14 \text{ lbs/(ft/sec)} \]2. **Formulating the Differential Equation:** - For mass-spring-damper systems: \( m \ddot{u}(t) + c \dot{u}(t) + ku(t) = 0 \). - Converting the mass weight to mass \( (m) \): \[ W = mg \implies 24 = m \cdot 32 \] \[ m = \frac{24}{32} = \frac{3}{4} \text{ slugs} \] - Plugging in the values: \[ \frac{3}{4} \ddot{u}(t) + 14 \dot{u}(t) + 144 u(t) = 0 \] - Simplifying the differential equation:

Given,Weight, W = 24 lbsSpring is stretches by, x = 2 inchesviscous force, fd =28 lbsmass has…

Answered: A mass weighting 24 lbs stretches a…

A mass weighting 24 lbs stretches a spring 2 inches. The mass is in a medium that exerts a viscous resistance of 28 lbs when the mass has a velocity of 2 ft/sec.

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Concept explainers

Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

Question

**Problem Statement:**

A mass weighing 24 lbs stretches a spring 2 inches. The mass is in a medium that exerts a viscous resistance of 28 lbs when the mass has a velocity of 2 ft/sec.

Suppose the object is displaced an additional 6 inches and released.

Find an equation for the object's displacement, \( u(t) \), in feet after \( t \) seconds.

\[ u(t) = \] (with a blank placeholder for the equation)

---

### Analysis and Problem-Solving Steps:

1. **Determine Constants:**
- **Spring Constant (k):**
The weight of the mass, \( W = 24 \) lbs, stretches the spring by 2 inches \( = \frac{2}{12} \) ft \( = \frac{1}{6} \) ft.
Using Hooke's Law: \( W = kx \), where \( x \) is the displacement,
\[ 24 = k \left( \frac{1}{6} \right) \]
Solving for \( k \):
\[ k = 24 \times 6 = 144 \text{ lbs/ft} \]

- **Damping Coefficient (c):**
The viscous resistance is \( 28 \) lbs at \( 2 \) ft/sec,
\[ c \cdot 2 = 28 \]
Solving for \( c \):
\[ c = \frac{28}{2} = 14 \text{ lbs/(ft/sec)} \]

2. **Formulating the Differential Equation:**
- For mass-spring-damper systems: \( m \ddot{u}(t) + c \dot{u}(t) + ku(t) = 0 \).
- Converting the mass weight to mass \( (m) \):
\[ W = mg \implies 24 = m \cdot 32 \]
\[ m = \frac{24}{32} = \frac{3}{4} \text{ slugs} \]

- Plugging in the values:
\[ \frac{3}{4} \ddot{u}(t) + 14 \dot{u}(t) + 144 u(t) = 0 \]

- Simplifying the differential equation:

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