Answer to Question #121355 in Calculus for Daniel

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

Natural length, $l_0 = 10 \ in = 0.254 \ m$

Stretching, $x = 1.5 \ in = 0.0381 \ m$

Weight, $m = 8 \ pounds = 3.632 \ kg$

Let spring constant is $k$ then $k= x\ mg = \frac{3.632∗10}{0.0381} =953.3 N m^{−1}$

a) Work done in stretching spring from normal to 14 inches

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

(b) Work done in stretching from 11 inches to 13 inches

$W = \frac{1}{2}k(l_2^{2} - l_1^{2}) = \frac{1}{2} \times953.3 \times((0.3302)^{2} - (0.2794)^{2})=14.76J$