Shared08 - Work Assignment / Work Notes.sagewsOpen in CoCalc
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Prerequisites:

  • Intro to Sage
  • Symbolic Integration

Work

The work WW that a constant force FF does when moving an object over distance dd is W=FdW=Fd.

The unit of work is the unit of force times the unit of distance. In SI units, force is measured in newtons (NN) and distance is measured in meters (mm), so work is measured in newton-meters, also called joules (JJ). In the British system, work is measured in foot-pounds.

If the force is not constant, then calculating work is more complicated. Suppose the force is given by a function F(x)F(x), where xx is the position of the object. We want to find the work done moving an object from x=ax=a to x=bx=b. So we divide the interval [a,b][a,b] into nn equal subintervals of width Δx\Delta x, and we choose a point xix_i from the ithi^{th} subinterval. If the subintervals are small, then the force on each subinterval is approximately the constant F(xi)F(x_i), so the work required to move across the ithi^{th} subinterval is approximately F(xi)ΔxF(x_i)\Delta x, and the total work is approximately i=1nF(xi)Δx\displaystyle\sum_{i=1}^n F(x_i)\Delta x. The actual work is the limit of this sum as nn\to\infty. Since this is a Riemann sum, the answer is:

W=abF(x)dx W=\int_a^bF(x)\, dx

Springs

One place where work arises is when stretching or compressing springs. To answer questions about springs, we need to know the relationship between force and distance.

Hooke's Law: The force required to hold a stretched or compressed spring a distance of xx units from its natural length is F(x)=kxF(x)=kx, where k>0k>0 is a constant (called the spring constant; this depends on the nature of the spring).

Example 1

A spring with natural length 5 m5\ m and spring constant k=30 N/mk=30\ N/m is stretched 10 m10\ m from its natural length. Find the work required.

Solution W=010F(x)dx=01030xdx=1500 JW=\int_0^{10}F(x)\, dx=\int_0^{10}30x\, dx=1500\ J

integral(30*x,x,0,10)
1500

Example 2

A spring has a natural length of 20 cm20\ cm. A 40 N40\ N force is required to stretch and hold the spring at a length of 30 cm30\ cm. How much work (in joules) is done in stretching the spring from 35 cm35\ cm to 38 cm38\ cm?

Solution In order to calculate the work, we must integrate the force function. That means we need to find kk for this particular spring. We will use joules as the unit of work, so we must make sure we use newtons for force and meters for distance.

We are told that 40 N40\ N of force is required to hold the spring 10 cm=0.1 m10\ cm = 0.1\ m from the natural length (302030-20). That means 40=k0.140=k\cdot0.1, or k=400k=400.

We want to move the spring from 35 cm35\ cm to 38 cm38\ cm, which is 0.15 m0.15\ m to 0.18 m0.18\ m beyond the natural length. Thus, the work is:

W=0.150.18400xdx=1.98 JW=\int_{0.15}^{0.18}400x\, dx=1.98\ J

integral(400*x,x,.15,.18)
1.9799999999999998

Example 3

We have a cable that weighs 2 lbs/ft attached to a bucket filled with coal that weighs 800 lbs. How much work (in foot-pounds) does it take to raise the bucket from the bottom of a 500 ft mine shaft?

Solution The force involved is the weight of the cable plus the weight of the bucket. The weight of cable depends on how much cable is left.

Let xx be the length of cable remaining. Then the force is F(x)=2x+800F(x)=2x+800.

Therefore, the work is W=05002x+800dx=650,000 ft-lb.W=\int_0^{500}2x+800\, dx=650,000\text{ ft-lb}.

integral(2*x+800,x,0,500)
650000

Example 4

A tank has the shape of an inverted circular cone with height 15 ft and base radius 4 ft. It is filled with water to a depth of 12 ft. Find the work needed to pump all the water to the top of the tank (the pump is floating on the water). Note: Water weighs 62.4 lbs/ft362.4\text{ lbs/ft}^3.

Solution We can't find a simple force function F(x)F(x), so this problem is more complicated than the previous examples.

Let xx be the vertical distance from the bottom of the tank (x=0x=0). The top of the tank is at x=15x=15, and the water occupies the interval from 00 to 1212.

Divide [0,12][0,12] into nn equal subintervals of width Δx\Delta x, and let xix_i be a point in the ithi^{th} subinterval.

For each subinterval we'll approximate the water with a cylinder of radius rir_i and height Δx\Delta x.

By similar triangles, rixi=415\displaystyle\frac{r_i}{x_i}=\frac{4}{15}, so ri=415xir_i=\frac{4}{15}x_i.

Here is a picture of a cross section of the tank:

Force is provided by the weight of the water, which is 62.4 lbs/ft362.4\text{ lbs/ft}^3 times the volume (in cubic feet).

The volume of water in the ithi^{th} subinterval is Vivolume of cylinder=πr2h=πri2Δx=π(415xi)2Δx.V_i\approx\text{volume of cylinder}=\pi r^2h=\pi r_i^2\Delta x=\pi \left(\frac{4}{15}x_i\right)^2\Delta x.

So the force (= weight of water) for the ithi^{th} subinterval is Fi=62.4Vi62.4π(415xi)2Δx=1664π375xi2Δx.F_i=62.4V_i\approx62.4\pi \left(\frac{4}{15}x_i\right)^2\Delta x=\frac{1664\pi}{375} x_i^2\Delta x.

The distance required to get the water from the ithi^{th} subinterval to the top of the tank is di15xid_i\approx 15-x_i, so the work required to pump the ithi^{th} subinterval is Wi=Fidi1664π375xi2Δx(15xi).W_i= F_id_i\approx\frac{1664\pi}{375} x_i^2\Delta x(15-x_i).

The total work reqired is Wi=1nWii=1n1664π375xi2(15xi)Δx\displaystyle W\approx\sum_{i=1}^{n}W_i\approx\sum_{i=1}^{n}\frac{1664\pi}{375} x_i^2(15-x_i)\Delta x.

To get the actual value of the work, we take the limit as nn\to\infty. Notice that the sum above is a Riemann sum, so W=limni=1n1664π375xi2(15xi)Δx=0121664π375x2(15x)dx48,178 ft-lb.W=\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1664\pi}{375} x_i^2(15-x_i)\Delta x=\int_0^{12}\frac{1664\pi}{375} x^2(15-x)\, dx\approx48,178\text{ ft-lb}.

integral(1664*pi/375*x^2*(15-x),x,0,12)
N(_)
1916928/125*pi 48177.6553780846