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2. A 2 kg mass stretches a spring 0.4 m. The spring constant is (g = 9.8- m and m kg. = N) s2 (a) 0.2 kg/m (b) 49 N/m (c) 0.8 kg . m (d) 160 N 3. A 24 lb weight stretches a spring 3 in. The spring constant is (а) 8 b/ft (b) 76 Įb/ft (c) 6 lb/ft (d) 96 lb/ft

2. A 2 kg mass stretches a spring 0.4 m. The spring constant is (g = 9.8- m and m kg. = N) s2 (a) 0.2 kg/m (b) 49 N/m (c) 0.8 kg . m (d) 160 N 3. A 24 lb weight stretches a spring 3 in. The spring constant is (а) 8 b/ft (b) 76 Įb/ft (c) 6 lb/ft (d) 96 lb/ft

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**2. A 2 kg mass stretches a spring 0.4 m. The spring constant is (g = 9.8 m/s² and kg·m/s² = N)** (a) 0.2 kg/m (b) 49 N/m (c) 0.8 kg·m (d) 160 N --- **3. A 24 lb weight stretches a spring 3 in. The spring constant is** (a) 8 lb/ft (b) 76 lb/ft (c) 6 lb/ft (d) 96 lb/ft --- **Explanation of Concepts:** The problems above relate to Hooke’s Law, which states that the force needed to extend or compress a spring by some distance \( x \) is proportional to that distance. This is often formulated as \( F = kx \), where \( k \) is the spring constant. **Units:** - In problem 2, the metric system is used with units of kilograms, meters, and Newtons (kg, m, N). - In problem 3, the imperial system is used with units of pounds and inches (lb, in). **Calculations:** 1. **For Problem 2:** - Weight \( W = mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \) - The spring constant \( k = \frac{F}{x} = \frac{19.6 \, \text{N}}{0.4 \, \text{m}} = 49 \, \text{N/m} \) 2. **For Problem 3:** - Note that 3 inches must be converted to feet: \( 3 \, \text{in} = 0.25 \, \text{ft} \) - Spring constant \( k = \frac{F}{x} = \frac{24 \, \text{lb}}{0.25 \, \text{ft}} = 96 \, \text{lb/ft} \)
Transcribed Image Text:**2. A 2 kg mass stretches a spring 0.4 m. The spring constant is (g = 9.8 m/s² and kg·m/s² = N)** (a) 0.2 kg/m (b) 49 N/m (c) 0.8 kg·m (d) 160 N --- **3. A 24 lb weight stretches a spring 3 in. The spring constant is** (a) 8 lb/ft (b) 76 lb/ft (c) 6 lb/ft (d) 96 lb/ft --- **Explanation of Concepts:** The problems above relate to Hooke’s Law, which states that the force needed to extend or compress a spring by some distance \( x \) is proportional to that distance. This is often formulated as \( F = kx \), where \( k \) is the spring constant. **Units:** - In problem 2, the metric system is used with units of kilograms, meters, and Newtons (kg, m, N). - In problem 3, the imperial system is used with units of pounds and inches (lb, in). **Calculations:** 1. **For Problem 2:** - Weight \( W = mg = 2 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 19.6 \, \text{N} \) - The spring constant \( k = \frac{F}{x} = \frac{19.6 \, \text{N}}{0.4 \, \text{m}} = 49 \, \text{N/m} \) 2. **For Problem 3:** - Note that 3 inches must be converted to feet: \( 3 \, \text{in} = 0.25 \, \text{ft} \) - Spring constant \( k = \frac{F}{x} = \frac{24 \, \text{lb}}{0.25 \, \text{ft}} = 96 \, \text{lb/ft} \)
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