Answer to Question #121241 in Calculus for CLEMENT

Question

A spring of natural length 10 inches stretches 1.5 inches under a
weight of 8 pounds. Find the work done in stretching the spring.
(a) from its natural length to a length of 14 inches
(b) from a length of 11 inches to a length of 13 inches.

Answer 1

Convert length to metre and mass into kilogram for simplicity.

natural length, $l_0 = 10in = 0.254m$

stretching, $x = 1.5in = 0.0381m$

weight, $m = 8pounds = 3.632kg$

let spring constant is $k$

then $k = \frac{mg}{x} = \frac{3.632*10}{0.0381} = 953.3 Nm^{-1}$

(i) work done in stretching spring from normal to 14 inches.

$W = \frac{1}{2}k(l_2^{2} - l_1^{2}) = \frac{1}{2}*953.3*((0.3556)^{2} - (0.254)^{2}) = 29.5 J$

(ii) work done in stretching from 11 inches to 13 inches

$W = \frac{1}{2}k(l_2^{2} - l_1^{2}) = \frac{1}{2}*953.3*((0.3302)^{2} - (0.2794)^{2})=14.76J$