Physics problems and workJon LeeJanuary 25, 2007Recall the following physics formulae:F = m · a (force = mass × acceleration)W = F · d (work = force × distance)m = ρ · V (mass = density × volume) .6.5.1 A particle is moved along the x-axis by a force that measures 10/(1 + x)2pounds at apoint x feet from the origin. Find the work done in moving the particle from the originto a distance of 9 ft.To set up the integral, divide the path of the particle into segments, and for eachsegment, let x denote its distance from the origin and ∆x denote its length. For eachsuch segment, the work done to move the particle across it is approximatelywork across a segment ≈ 10/(1 + x2) lb| {z }force· ∆x ft|{z}distance(W = F · d)= 10/(1 + x)2· ∆x lb·ft .Summing over all our segments from x = 0 to x = 9, this istotal work across path ≈X10/(1 + x)2lb·ft .As our segments become smaller, we obtain the integraltotal work across path =Z9010/(1 + x2) · ∆x dx lb·ft= 10Z1011/u2du lb·ft= 10 · (−1/u)|1u=10 lb·ft= 10(1 − 1/10) lb·ft= 9 lb·ft .16.5.11 A cable that weighs 2 lb/ft is used to lift 800 lb of coal up a mineshaft 500 ft deep. Findthe work done.First, we find the work done on the cable itself. To set up the integral, divide thecable into segments, and for each segment, let x denote its distance from the top ofthe mineshaft and ∆x denote its length. For such a segment, the work done to moveit to the top is approximatelywork on a cable segment ≈force densityz }| {(2 lb/ft)lengthz }| {(∆x ft)| {z }force· x ft|{z}distance(W = F · d)= 2 · x · ∆x lb·ft .Summing over all our segments from x = 0 to x = 500, this istotal work on the cable ≈X2 · x · ∆x lb·ft .As our segments become smaller, we obtain the integraltotal work on the cable =Z50002x dx lb·ft= x2500x=0lb·ft= 250000 lb·ft .It remains to calculate the work done on the coal. Since all the coal is concentrated atthe end of the rope, there’s no need to approximate our integral using slices. Thus, wejust havework done on coal = 800 lb| {z }force· 500 ft| {z }distance(W = F · d)= 400000 lb·ft .Thus, our total work done istotal work = 250000 lb·ft| {z }on cable+ 400000 lb·ft| {z }on coal= 650000 lb·ft .6.5.15 An aquarium 2 m long, 1 m wide, and 1 m deep is full of water. Find the work neededto pump half of the water out of the aquarium. (Use the fact that the density of wateris 1000 kg/m3.)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 denote2its thickness. (We choose horizontal slices because we want each drop of water in agiven slic e to be the same distance from the top of the tank.) Using the formulae atthe beginning of this handout, we see that the work taken to pump such a slice out ofthe tank iswork 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 becomeswork for a slice ≈forcez }| {massz }| {(1000 kg/m3)| {z }density(2∆x m3)| {z }volume(9.8 m/s2)| {z }gravity(x m)|{z}distance= (1000)(9.8)(2)x · ∆x (kg · m/s2) · m= (1000)(9.8)(2)x · ∆x N · m= (1000)(9.8)(2)x · ∆x J .Summing over our slices, this istotal 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 distancex = 0 to x = 1/2. As we refine our slices, this becomes the integraltotal work =Z1/20(1000)(9.8)(2)x dx J= (1000)(9.8)(2)Z1/20x dx J= (1000)(9.8)(2)(1/8) J= 2450 J