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

$4640 J$
We integrate in order to find the work done: $$W=W_1+W_2 \\= \int_{0}^{100}(100) dx+ \int_{0}^{40} (0.8) (40-x) dx \\=[100 x]_0^{40} +0.8 \times [40 x-\dfrac{x^2}{2}]_0^{40} \\=4000+ \dfrac{(0.8) (1600)}{2} \\=4000+640 \\ =4640 J$$