if a spring has a spring constant of 3 b/inch, how far would it stretch if 10 lb were hung from its end?

Explanation

To calculate how far the spring would stretch when a 10 lb weight is hung from it, we can use Hooke's Law: 

$F = k \cdot x$

where

• \(F\) is the force applied (10 lb),
• \(k\) is the spring constant (3 lb/inch),
• \(x\) is the displacement of stretch in inches.

Rearranging the formula to solve for \(x\):

$x = \frac{F}{k}$

Substituting the vaues:

$x = \frac{10 \text{ lb}}{3 \text{ lb/inch}} = \frac{10}{3} ≈ 3.33 \text{ inches} $

So, the spring would stretch approximately 3.33 inches.