Chapter 6: Applications of Definite Integrals - Section 6.5 - Work and Fluid Forces - Exercises 6.5 - Page 348: 9

$72900 ft-lb$

Hooke's Law states that $F=k x$ Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$ This implies that $W=\int_0^{(180)} (4.5)(x) dx$ Use formula: $\int x^n=\dfrac{x^{n+1}}{n+1}+C$ Then $W=(4.5)(\dfrac{x^2}{2})]_0^{180}=2.25[(180)^2-0]=(2.25)(32400)=72900 ft-lb$