Math Problem Statement

A force of 320 lb is required to hold a spring stretched 5 ft beyond its natural length. How much work W is done in stretching it from its natural length to 3 inches beyond its natural length?

Solution

To solve this problem, we first need to understand the relationship between force, spring constant, and displacement in Hooke's Law, which states:

F=kxF = kx

where:

  • FF is the force applied to the spring,
  • kk is the spring constant,
  • xx is the displacement from the natural length.

Step 1: Find the Spring Constant kk

Given that a force of 320320 lb is required to stretch the spring 55 ft (or 6060 inches) beyond its natural length, we can find the spring constant kk using Hooke's Law:

320=k×60320 = k \times 60

Solving for kk:

k=32060=163lb/ink = \frac{320}{60} = \frac{16}{3} \, \text{lb/in}

Step 2: Calculate the Work Done WW

The work done in stretching a spring is given by the integral of the force over the displacement:

W=0xFdx=0xkxdxW = \int_{0}^{x} F \, dx = \int_{0}^{x} kx \, dx

where xx is the displacement from the natural length. In this case, we need to calculate the work done in stretching the spring from its natural length to 33 inches:

W=03163xdxW = \int_{0}^{3} \frac{16}{3} x \, dx

Evaluating the integral:

W=16303xdx=163[x22]03W = \frac{16}{3} \int_{0}^{3} x \, dx = \frac{16}{3} \left[\frac{x^2}{2}\right]_{0}^{3}

W=163×322=163×92=16×96=24ft-lbW = \frac{16}{3} \times \frac{3^2}{2} = \frac{16}{3} \times \frac{9}{2} = \frac{16 \times 9}{6} = 24 \, \text{ft-lb}

Therefore, the work done in stretching the spring from its natural length to 33 inches beyond its natural length is 24 ft-lb.

Would you like any further details or have any questions? Here are some related problems you might find interesting:

  1. How much work is done in stretching the spring from 3 inches to 5 feet beyond its natural length?
  2. What would be the spring constant if the force required was 400 lb for the same stretch?
  3. How does the work change if the spring is stretched only 2 inches instead of 3?
  4. What is the potential energy stored in the spring when stretched 3 inches beyond its natural length?
  5. How much force is required to stretch the spring 10 inches beyond its natural length?

Tip: Always make sure units are consistent when performing calculations in physics, especially with mixed units like feet and inches!

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Math Problem Analysis

Mathematical Concepts

Physics
Hooke's Law
Work-Energy Principle
Integration

Formulas

Hooke's Law: F = kx
Work Done by a Spring: W = ∫(k * x) dx

Theorems

Hooke's Law
Work-Energy Theorem

Suitable Grade Level

Grades 11-12 (Advanced High School Physics)