# BC Calculus More Work Problems

• Class: 5H: BC Calculus
• Author: Peter Atlas
• Text: Calculus Finney, Demana, Waits, Kennedy

CALCULATOR INACTIVE

1. When a spring is expanded 1 foot from its natural position and held fixed, the force necessary to hold it is 30 pounds. Find the work required to stretch the spring from 3 to 6 feet.
2. Solution $$\displaystyle 30 = 1 \cdot k \implies k = 30. \int_3^6 30x \, dx = \left. 15x^2 \right|_3^6 = 15 \left( 6^2 - 3^2 \right)$$ foot pounds
3. If 6 foot-pounds of work are required to compress a spring 1 foot from its natural length, find the work necessary to compress the spring 1 extra foot.
4. Solution $$\displaystyle 6 = \int_0^1 kx \, dx \implies \left. \frac{kx^2}{2} \right|_0^1 = \frac{k}{2}$$, so $$k = 12$$. $$\displaystyle \int_1^2 12x \, dx \implies \left. \frac{kx^2}{2} \right|_1^2$$ = $$\displaystyle \left( \frac{12 \cdot 2^2}{2} \right) - \left( \frac{12 \cdot 1^2}{2} \right) foot pounds.$$.
5. If 6 foot-pounds of work are required to compress a spring from its natural length of 10 feet to a length of 9 feet, find the work necessary to stretch the spring from its natural length to a length of 12 feet.
6. Solution $$\displaystyle 6 = \int_0^1 kx \, dx \implies \left. \frac{kx^2}{2} \right|_0^1 = \frac{k}{2}$$, so $$k = 12$$. $$\displaystyle \int_0^{-2} 12x \, dx \implies \left. \frac{kx^2}{2} \right|_0^{-2}$$ = $$\displaystyle \left( \frac{12 \cdot (-2)^2}{2} \right)$$ foot pounds..
7. A cylindrical drum of radius 4 feet and height 10 feet is cut in half and layed on its side so that the bases are now vertical semicircles, 10 feet apart as in the figure below. The drum is used as an open top tank to store gasoline, despite the myriad state and federal laws prohibiting such a practice. If the tank is full, find the work necessary to pump the gasoline to the top of the tank. Assume gasoline weighs 42 pounds per cubic foot.
8. Solution I'll place the origin at the center of the semi circular base of the cylinder, with the drum extending below the x-axis. Then the equation of the drum is $$\displaystyle x^2 + y^2 = 16 \implies x = \sqrt{16 - y^2}$$. $$\displaystyle \int_{y = -4}^{y = 0} 2x \cdot 10 \cdot 42 \cdot (0 - y) \, dy$$ = $$\displaystyle -840 \int_{-4}^{0} y \sqrt{16 - y^2} \, dy$$ Let $$\displaystyle u = y^2$$ Then $$\displaystyle \frac{du}{2} = y dy, u(0) = 0, u(-4) = 16.$$ In $$u$$, the integral becomes $$\displaystyle -420 \int_{16}^{0} (16 - u)^{\frac{1}{2}} \, du$$ = $$\displaystyle 420 \int_{0}^{16} (16 - u)^{\frac{1}{2}} \, du$$ = $$\displaystyle 420 \left. \left( -\frac{2}{3} (16 - u)^{\frac{3}{2}} \right) \right|_{16}^{0}$$ = $$\displaystyle 420 \left( -\frac{2}{3} (16 - 16)^{\frac{3}{2}} \right) - 420 \left( -\frac{2}{3} (16 - 0)^{\frac{3}{2}} \right)$$ foot pounds.
9. If the tank in the figure above is inverted so that the rectangular surface is on the ground and filled with water which weighs $$w$$ lbs/cubic foot, find the work necessary to pump all the water out a spout at the top.
10. Solution Now I'm going to assume the origin is still the center of the semi circular base, but this time extending above the x-axis. Then the equation of the drum is $$\displaystyle x^2 + y^2 = 16 \implies x = \sqrt{16 - y^2}$$. $$\displaystyle \int_{y = 0}^{y = 4} 2x \cdot 10 \cdot w \cdot (4 - y) \, dy$$ = $$\displaystyle 20w \int_{0}^{4} (4 - y) \sqrt{16 - y^2} \, dy$$ I'll break this up into two integrals: $$\displaystyle 80w \int_{0}^{4} \sqrt{16 - y^2} \, dy$$ - $$\displaystyle 20w \int_{0}^{4} y \sqrt{16 - y^2} \, dy$$. The first one I recognize as $$80w$$ times the area of a quarter circle with radius 4 = $$\displaystyle 80w \cdot \frac{ \pi 4^2}{4} = 320w \pi$$. For the second one, let $$\displaystyle u = y^2$$ Then $$\displaystyle \frac{du}{2} = y dy, u(4) = 16, u(0) = 0.$$ In $$u$$, the integral becomes $$- \displaystyle 10w \int_{0}^{16} (16 - u)^{\frac{1}{2}} \, du$$ = $$\displaystyle -10w \cdot \frac{2}{3} \cdot 64.$$ So the final answer is $$320w \pi - 10w \cdot \frac{2}{3} \cdot 64$$ foot pounds.
11. An elevator in the Empire State Building weighs 1600 pounds. Find the work required to raise the elevator from the ground level to the 102nd story, some 1200 feet above the ground.
12. Solution No calculus here. Nothing is varying. 1600 lbs times 1200 feet = 1,920,000 foot pounds
13. A bucket of cement weighing 200 pounds is hoisted by means of a windlass from the ground to the tenth story of an office building, 80 feet above the gound. If the weight of the rope used is negligible, find the work required to make the lift.
14. Solution No calculus here either. 200 lbs times 80 feet = 16,000 foot pounds
15. Assume that a chain weighing 1 pound per foot is used in the last exercise, instead of the lightweight rope. Find the work required to make the lift.
16. Solution $$\displaystyle 16,000 + \int_0^{80} (80 - x) \, dx$$ = $$\displaystyle 16,000 + \left. \left( 80x - \frac{x^2}{2} \right) \right|_0^{80}$$ = $$\displaystyle 16,000 + 80 \cdot 80 - \frac{80^2}{2}$$ foot pounds.
17. A container weighing 1 pound is lifted vertically at the rate of 2 feet per second. Water is leaking out of the container at the rate of $$\frac{1}{2}$$ pound per second. If the initial weight of the water and container is 20 pounds, how much work is done in raising the container 10 feet?
18. Solution The container weighs 1 pound and is lifted 10 feet, so the work for the container is 10 * 1 = 10 foot pounds. For the water, since the container and water starts weighing 20 pounds, the water must start weighing 19 pounds. The weight of the water as a function of time is $$w(t) = 19 - \frac{1}{2}t$$. Now, the bounds have to be in $$t$$, so $$10 = 2t \implies t = 5.$$ So total work = $$\displaystyle 10 + \int_0^{5} \left( 19 - \frac{1}{2}t \right) \, dy$$. Since $$\displaystyle y = 2t, dy = 2dt$$ then $$\displaystyle 10 + 2 \int_0^{5} \left( 19 - \frac{1}{2}t \right) \, dt$$ = $$\displaystyle 10 + \left. 2 \left( 19t - \frac{1}{4}t^2 \right) \right|_0^5$$ = $$\displaystyle 10 + 2 \left( 19 \cdot 5 - \frac{1}{4}5^2 \right)$$
19. A bucket containing water is raised vertically at the rate of 2 feet per second. Water is leaking out of the rate of $$\frac{1}{2}$$ pound per second. If the bucket weighs 1 pound and initially contains 20 pounds of water, determine the amount of work required to raise the bucket until it is empty.
20. Solution The bucket weighs 1 pound and is lifted until it's empty. That means $$0 = 20 - \frac{1}{2}t \implies 40 = t$$, so $$y = 2 \cdot 40 = 80$$ feet, so the work for the bucket is 80 * 1 = 80 foot pounds. For the water, since the water starts weighing 20 pounds, the weight of the water as a function of time is $$w(t) = 20 - \frac{1}{2}t$$. Now, the bounds have to be in $$t$$, so $$0 = 20 - \frac{1}{2}t \implies 40 = t.$$ So total work = $$\displaystyle 10 + \int_0^{40} \left( 20 - \frac{1}{2}t \right) \, dy$$. Since $$\displaystyle y = 2t, dy = 2dt$$ then $$\displaystyle 80 + 2 \int_0^{40} \left( 20 - \frac{1}{2}t \right) \, dt$$ = $$\displaystyle 80 + \left. 2 \left( 20t - \frac{1}{4}t^2 \right) \right|_0^{40}$$ = $$\displaystyle 80 + 2 \left( 20 \cdot 40 - \frac{1}{4}40^2 \right)$$ foot pounds.
21. The great pyramid of Cheops has square cross sections and was (approximately) 482 feet high and 754 feet per side at the base. If the rock used to build the pyramid weighed 150 pounds per cubic foot, find the work required to lift the rock into place as the pyramid was built.
22. Solution Put the origin at the center of the square base, and partition the height y. Calling the height of the partition $$y$$ and the edge of the square base $$e$$, inscribing similar triangles into the pyramid, we get $$\displaystyle \frac{482 - y}{482} = \frac{e}{754} \implies e = \frac{754(482 - y)}{482}.$$ We want $$\displaystyle 150 \int_{y = 0}^{y=482} e^2 y \, dy$$ = $$\displaystyle 150 \int_{0}^{482} \left( \frac{754(482 - y)}{482} \right) ^2 y \, dy$$ Harumph. This is silly to do without a calculator. From my calculator, I get approx. $$1.6509 x 10^{12}$$.
23. A ball weighing 0.2 pounds is thrown vertically upward. Its height after $$t$$ seconds is given by $$h(t) = 6 + 8t - 16t^2$$ until the ball strikes the ground again. Find the work done on the ball by gravity while the ball descends from its maximum height to the ground.
24. Solution After a lot of fumbling about, I realized this isn't a Calculus question. This is an object of constant weight (0.2 lbs) travelling a constant distance: from the vertex of $$h(t) = 6 + 8t - 16t^2$$ to the t-intercept. The vertex occurs at $$t = -\frac{B}{2A} \implies \left( \frac{1}{4}, 7 \right) . 0.2 \cdot 7 = 1.4$$ foot pounds.