A 24 Ib weight stretches a spring 3 in. The spring constant is (a) 8 Ib/ft (b) 76 Ib/ft (c) 6 Ib/ft (d) 96 Ib/ft
A 24 Ib weight stretches a spring 3 in. The spring constant is (a) 8 Ib/ft (b) 76 Ib/ft (c) 6 Ib/ft (d) 96 Ib/ft
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![**Problem Statement:**
A 24 lb weight stretches a spring 3 inches. The spring constant is:
(a) 8 lb/ft
(b) 76 lb/ft
(c) 6 lb/ft
(d) 96 lb/ft
**Explanation of the Problem:**
This problem requires finding the spring constant (k) given the force (F) applied to the spring and the displacement (x) caused by the force. Use Hooke’s Law, which states:
\[ F = kx \]
Where:
- \( F \) is the force applied in pounds (lb),
- \( k \) is the spring constant in pounds per foot (lb/ft),
- \( x \) is the displacement in feet (ft).
**Conversion:**
First, convert the displacement from inches to feet:
\[ 3 \text{ inches} = \frac{3}{12} \text{ feet} = 0.25 \text{ feet} \]
**Calculation:**
Using Hooke’s Law:
\[ F = kx \]
\[ 24 = k \times 0.25 \]
Solving for \( k \):
\[ k = \frac{24}{0.25} = 96 \text{ lb/ft} \]
Therefore, the correct answer is:
(d) 96 lb/ft](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcc2e6b79-1f48-4c40-b6f8-4e5fcdb801ee%2Fae920ba1-7dc8-4b0e-8333-2d1390582bab%2Fz2e6upb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A 24 lb weight stretches a spring 3 inches. The spring constant is:
(a) 8 lb/ft
(b) 76 lb/ft
(c) 6 lb/ft
(d) 96 lb/ft
**Explanation of the Problem:**
This problem requires finding the spring constant (k) given the force (F) applied to the spring and the displacement (x) caused by the force. Use Hooke’s Law, which states:
\[ F = kx \]
Where:
- \( F \) is the force applied in pounds (lb),
- \( k \) is the spring constant in pounds per foot (lb/ft),
- \( x \) is the displacement in feet (ft).
**Conversion:**
First, convert the displacement from inches to feet:
\[ 3 \text{ inches} = \frac{3}{12} \text{ feet} = 0.25 \text{ feet} \]
**Calculation:**
Using Hooke’s Law:
\[ F = kx \]
\[ 24 = k \times 0.25 \]
Solving for \( k \):
\[ k = \frac{24}{0.25} = 96 \text{ lb/ft} \]
Therefore, the correct answer is:
(d) 96 lb/ft
Expert Solution
Step 1
In equilibrium, the net force acting on an object will be 0.
If a spring is streched by x distance, then the restoring Force acting will be where k is the spring constant.
Here the downward weight of the object stretches the spring until it reaches equilibrium where the upward force due to spring balances the downward weight.
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