Question

An aquarium 7 m long, 1 m wide, and 1 m deep is full of water. find the work needed to pump half of the water out of the aquarium. (use 9.8 m/s2 for g and the fact that the density of water is 1000 kg/m3.) show how to approximate the required work by a riemann sum. (let x be the height in meters below the top of the tank. enter xi* as xi.) lim n → ∞ n i = 1 δx express the work as an integral. 0 dx evaluate the integral. j

Answer 1

The work needed to pump out half the water from the given aquarium is 17,150 Joules. This is determined by applying physical principles to calculate the mass of the water, the effect of gravity, and using an integral to find the total work done against gravity.

Step-by-step explanation:

The student is asking about the work required to pump half of the water out of an aquarium with dimensions 7 m long, 1 m wide, and 1 m deep using physics concepts involving work, force, and Riemann sums. To approach this problem, we must consider the work done against gravity to move the water from its initial position to the top of the aquarium. The density of water (ρ) is 1000 kg/m³, and the acceleration due to gravity (g) is 9.8 m/s².

First, calculate the volume of water to be pumped out, which is half the aquarium volume: V = ½ × 7 m × 1 m × 1 m = 3.5 m³. Convert this volume to mass using the density of water, m = ρV = 1000 kg/m³ × 3.5 m³ = 3500 kg.

The work done to pump out half the water can be calculated using the concept of the center of mass of the water being lifted, which is at a height h/2 from the top of the water when the tank is half full, where h is the depth of the tank. Therefore, the work is W = mgh/2 = 3500 kg × 9.8 m/s² × 0.5 m = 17150 J.

To approximate the required work using a Riemann sum, consider the small amount of work to lift a thin layer of water δx from a depth x to the top of the tank, dW = ρgAdx(x), where A is the area of the tank's surface. We set up the integral ∫ W = ρgA ∫ xdx from 0 to h/2, and find the limit as the number of partitions goes to infinity. The integration gives us the same work W = 17150 J.

Answer 2

Answer: To set up the integral, we divide the upper half of the aquarium into horizontal slices, and for each slice, let x denote its distance from the top of the tank and ∆x denote 2 its thickness. (We choose horizontal slices because we want each drop of water in a given slice to be the same distance from the top of the tank.) Using the formulae at the beginning of this handout, we see that the work taken to pump such a slice out of the tank is work for a slice = W = F · d = (m · a) · d = (ρ · V ) · a · d . Since the length, width and thickness of the slice are given by 2 m, 1 m and ∆x m, respectively, its volume is 2 · 1 · ∆x m3 = 2∆x m3 . Thus, the equation above becomes work for a slice ≈ force z }| { mass z }| { (1000 kg/m 3 ) | {z } density (2∆x m 3 ) | {z } volume (9.8 m/s 2 ) | {z } gravity (x m) | {z } distance = (1000)(9.8)(2)x · ∆x (kg · m/s 2 ) · m = (1000)(9.8)(2)x · ∆x N · m = (1000)(9.8)(2)x · ∆x J . Summing over our slices, this is total work for top half of aquarium ≈ X(1000)(9.8)(2)x · ∆x J , where the sum is over the slices in the top half of the aquarium; that is, from distance x = 0 to x = 1/2. As we refine our slices, this becomes the integral total work = Z 1/2 0 (1000)(9.8)(2)x dx J = (1000)(9.8)(2) Z 1/2 0 x dx J = (1000)(9.8)(2)(1/8) J = 2450 J .