Math Problem Statement

A force of 3 pounds compresses a 15 inch spring a total of 3 inches. How much work (in ft-lbs) is done in compressing the spring 10 inches?

Solution

To solve this problem, we need to use Hooke's Law for springs and the work done to compress a spring. Hooke's Law states that the force required to compress or extend a spring is proportional to the displacement from its natural length:

F=kxF = kx

Where:

  • FF is the force applied,
  • kk is the spring constant (stiffness),
  • xx is the displacement (amount the spring is compressed or stretched).

Step 1: Find the Spring Constant kk

We are given:

  • A force F=3F = 3 pounds,
  • A compression distance x=3x = 3 inches.

Using Hooke’s Law:

3=k×33 = k \times 3

Solving for kk:

k=33=1 pound per inchk = \frac{3}{3} = 1 \text{ pound per inch}

Step 2: Set Up the Work Formula

The work done on a spring is given by the formula:

W=0xFdx=0xkxdxW = \int_0^x F \, dx = \int_0^x kx \, dx

Where:

  • WW is the work done,
  • xx is the final displacement.

For a spring with spring constant kk, the work to compress it from 00 to xx is:

W=12kx2W = \frac{1}{2} k x^2

Step 3: Calculate the Work for a 10-Inch Compression

We are asked to find the work required to compress the spring by 10 inches. Using the work formula:

W=12×1×(102)=12×100=50 inch-poundsW = \frac{1}{2} \times 1 \times (10^2) = \frac{1}{2} \times 100 = 50 \text{ inch-pounds}

Step 4: Convert Inch-Pounds to Foot-Pounds

Since there are 12 inches in a foot, we convert 50 inch-pounds to foot-pounds:

W=50124.17 foot-poundsW = \frac{50}{12} \approx 4.17 \text{ foot-pounds}

Thus, the work done in compressing the spring by 10 inches is approximately 4.17 foot-pounds.


Would you like more details on any part of the solution or have any further questions?

Related Questions:

  1. What is Hooke’s Law, and how is it applied in real-world examples?
  2. How does the spring constant affect the amount of work done in compressing or stretching a spring?
  3. What is the formula for the potential energy stored in a compressed spring?
  4. How do you convert between different units of work, such as inch-pounds and foot-pounds?
  5. What is the relationship between force and displacement for non-linear systems?

Tip: Always ensure unit consistency, especially when switching between systems like inches to feet!

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Math Problem Analysis

Mathematical Concepts

Physics
Hooke's Law
Work-Energy Principle

Formulas

Hooke's Law: F = kx
Work Formula for Springs: W = (1/2)kx^2
Unit Conversion: 1 foot = 12 inches

Theorems

Hooke's Law
Work-Energy Theorem

Suitable Grade Level

Grades 10-12