## University Calculus: Early Transcendentals (3rd Edition)

$1125$ N-m
We know that Hooke's Law states that $F=k x$ Then, $k=\dfrac{F}{x}=\dfrac{90}{01}=90$ To calculate the work done, the limits for $x$ will be from $0$ to $5$ such that: $W=\int_0^{5} 90 x dx= 90[ \dfrac{x^2}{2}]_0^{5}$ or, $W=45[ (5)^2-0] =1125$ N-m