9.5: Work
- Page ID
- 149527
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)A fundamental concept in classical physics is work: If an object is moved in a straight line against a force \(F\) for a distance \(s\) the work done is \(W=Fs\).
How much work is done in lifting a \(10\) pound weight vertically a distance of \(5\) feet?
Solution
The force due to gravity on a \(10\) pound weight is \(10\) pounds at the surface of the earth, and it does not change appreciably over \(5\) feet. The work done is \(W=10\times 5=50\) foot-pounds.
In reality few situations are so simple. The force might not be constant over the range of motion, as in the next example.
How much work is done in lifting a 10 pound weight from the surface of the earth to a height of 100 miles above the surface? (N.B. Height only, but not into orbit as well, which would take significantly more energy to accelerate to the necessary speed besides that required to simply achieve the height.)
Solution
Over \(100\) vertical miles the force due to gravity does change significantly, so we need to take this into account. The gravitational force exerted on a \(10\) pound mass at a distance \(r\) from the center of the earth is \(F=k/r^2\), and by definition it is \(10\) pounds weight when \(r\) is the radius of the earth (assumed spherical), which allows the constant \(k\) to be determined.
How can we approximate the work done? We divide the path from the surface to the final height into \(n\) small subpaths. On each subpath the force due to gravity is roughly constant, with value \(k/r_i^2\) at distance \(r_i\). The work to raise the object from \(r_i\) to \(r_{i+1}\) is thus approximately \((k/r_i^2)\Delta r\) and the total work is approximately \[\sum_{i=0}^{n-1} {k\over r_i^2}\Delta r,\nonumber\] or in the limit \[W=\int_{r_0}^{r_1} {k\over r^2}\,dr,\nonumber\] where \(r_0\) is the radius of the earth and \(r_1\) is \(r_0\) plus \(100\) miles. The work is \[W=\int_{r_0}^{r_1} {k\over r^2}\,dr= -\left.{k\over r}\right|_{r_0}^{r_1}=-{k\over r_1}+{k\over r_0}.\nonumber\] Assuming the earth to be spherical with radius \(3963\) miles, we have \(r_0=3963\times 5280\) and \(r_1=4063\times 5280\), both in feet. The force on the \(10\) pound weight at the surface of the earth is \(10\) pounds, so \(10=k/(3963\times 5280)^2\), giving \(k\approx 4.378\times 10^{15}\) approximately. Then \[-{k\over r_1}+{k\over r_0} = 10\times 3963^2\times 5280\left(\dfrac{1}{3963}-\dfrac{1}{4063}\right)\approx 5150000\;\text{foot-pounds}\text{, approximately.}\nonumber\] Note that if the force due to gravity is assumed to be \(10\) pounds over the whole distance, the work done would be calculated as \(10(r_1-r_0)=10\times 100\times 5280=5280000\) foot-pounds, which is somewhat higher since that calculation does not account for the weakening of the gravitational force.
How much work is done in lifting a \(10\) kilogram object from the surface of the earth to a distance \(D\) from the center of the earth?
Solution
This is the same problem as before in different units, and we are not specifying a value for \(D\). As before, \[W=\int_{r_0}^{D} {k\over r^2}\,dr= -\left.{k\over r}\right|_{r_0}^{D}=-{k\over D}+{k\over r_0}.\nonumber\] While "weight in pounds" is a measure of force, "weight in kilograms" is a measure of mass. To discover what force is acting upon a mass in a given situation, we need to use Newton's law \(F=ma\). At the surface of the earth the acceleration due to gravity is approximately \(9.8\) meters per second squared, (or \(9.8\) meters per second per second), so the gravitational force downwards on a \(10\) kilogram object is \(F=10\times 9.8=98\). The units of force here are "kilogram-meters per second squared" or "kg m/s\(^2\)", also known as Newtons (N), so \(F=98\) N. And if we take the radius of the (spherical) earth to be approximately \(6378.1\) kilometers or \(6378100\) meters, then the problem proceeds as before. From \(F=k/r^2\) we compute \(k\): \(98=k/6378100^2\), so \(k= 3.986655642\times 10^{15}\). Then the work is: \[W=-{k\over D}+6.250538000\times 10^8\quad\text{Newton-meters.}\nonumber\] As \(D\) increases \(W\) of course gets larger, since the quantity being subtracted, \(-k/D\), gets smaller. But note that the work \(W\) will never exceed \(6.250538000\times 10^8\), and in fact will approach this value as \(D\) gets larger. In short, with a finite amount of work, namely \(6.250538000\times 10^8\) N-m, we can lift the \(10\) kilogram object as far as we wish from earth; we might also say that this is the work required to lift the object to infinity.
Next is an example in which the force is constant, but there are many objects moving different distances.
Suppose that a water tank is shaped like a right circular cone with the tip at the bottom, and has height \(10\) meters and radius \(2\) meters at the top. If the tank is full, how much work is required to pump all the water out over the top?
Solution
Here we have a large number of atoms of water that must be lifted different distances to get to the top of the tank. Fortunately, we don't really have to deal with individual atoms---we can consider all the atoms at a given depth together.
To approximate the work, we can divide the water in the tank into horizontal sections, approximate the volume of water in a section by a thin disk, and compute the amount of work required to lift each disk to the top of the tank. As usual, we take the limit as the sections get thinner and thinner to get the total work.
Figure \(\PageIndex{1}\). A conical water tank.
At depth \(h\) the circular cross-section through the tank has radius \(r=(10-h)/5\), by similar triangles, and therefore its area is \(\pi(10-h)^2/25\) square meters. A section of the tank at depth \(h\) with thickness \(\Delta h\) thus has volume approximately \((\pi(10-h)^2/25)\times\Delta h\), and so it contains \(\sigma\times(\pi(10-h)^2/25)\Delta h\) kilograms of water, where \(\sigma\) is the density of water in kilograms per cubic meter; \(\sigma\approx 1000\). If the acceleration due to gravity is \(g\approx 9.8 m/s^2,\) the weight of this much water is \(g\times\sigma\pi((10-h)^2/25)\Delta h\) Newtons, and finally, this section of water must be lifted a distance \(h\), which requires \(h\times g\sigma\pi((10-h)^2/25)\Delta h\) Newton-meters of work. The total work needed to empty the tank out through the top is therefore \[W=\dfrac{g\sigma\pi}{25}\int_0^{10}h(10-h)^2\,dh = \dfrac{100 g\sigma\pi}{3}\;\hbox{Newton-meters,}\nonumber\] i.e. about \(1026000\) Newton-meters, to some (unspecified) level of accuracy.
A spring has a "natural length", which is its length if nothing is stretching or compressing it. If the spring is either stretched or compressed the spring provides an opposing force; according to Hooke's Law the magnitude of this force is proportional to the distance the spring has been stretched or compressed: \(F=kx\). The constant of proportionality, \(k\), of course depends on the spring. Note that \(x\) here represents the change in length from the natural length.
Suppose \(k=5\) for a given spring that has a natural length of \(0.1\) meters. Suppose a force is applied that compresses the spring to length \(0.08\). What is the magnitude of the force?
Solution
Assuming that the constant \(k\) has appropriate dimensions (namely, kg/sec\(^2\)), the force is \(5(0.1-0.08)=5(0.02)=0.1\) Newtons.
How much work is done in compressing the spring in the previous example from its natural length to \(0.08\) meters? From \(0.08\) meters to \(0.05\) meters? How much work is done to stretch the spring from \(0.1\) meters to \(0.15\) meters?
Solution
We can approximate the work by dividing the distance that the spring is compressed (or stretched) into small subintervals. Then the force exerted by the spring is approximately constant over each subinterval, so the work required to compress the spring from \(x_i\) to \(x_{i+1}\) is approximately \(5(x_i-0.1)\Delta x\). The total work is approximately \[\sum_{i=0}^{n-1} 5(x_i-0.1)\Delta x,\nonumber\] and in the limit \[W=\int_{0.1}^{0.08}5(x-0.1)\,dx=\left.{5(x-0.1)^2\over2}\right|_{0.1}^{0.08}= {5(0.08-0.1)^2\over2}-{5(0.1-0.1)^2\over2}={1\over1000}\,\text{N-m}.\nonumber\]
The other values we seek simply use different limits. To compress the spring from \(0.08\) meters to \(0.05\) meters takes \[W=\int_{0.08}^{0.05}5(x-0.1)\,dx = \left.{5(x-0.1)^2\over 2}\right|_{0.08}^{0.05} = {5(0.05-0.1)^2\over 2}-{5(0.08-0.1)^2\over 2}={21\over 4000}\;\text{N-m,}\nonumber\] and to stretch the spring from \(0.1\) meters to \(0.15\) meters requires \[W=\int_{0.1}^{0.15} 5(x-0.1)\,dx = \left.{5(x-0.1)^2\over 2}\right|_{0.1}^{0.15} = {5(0.15-0.1)^2\over 2}-{5(0.1-0.1)^2\over 2} = {1\over 160}\;\text{N-m.}\nonumber\]
Exercises \(\PageIndex{}\)
How much work is done in lifting a \(100\) kilogram mass from the surface of the earth to a height of \(35,786\) kilometers above the surface of the earth? Assume the same values for the spherical earth radius and the acceleration due to earth's gravity as in Example \(\PageIndex{3}\). An answer correct to two significant figures is sufficient.
- Answer
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\(\approx 5.3\times 10^9\) Newton-meters
How much work is done in lifting a \(100\) kilogram mass from a distance \(1000\) kilometers above the surface of the earth to a height of \(35,786\) kilometers above the surface of the earth?
- Answer
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\(\approx 4.5\times 10^9\) Newton-meters
A water tank has the shape of an upright cylinder with radius \(1\) meter and height \(10\) meters. If the depth of the water is \(5\) meters, how much work is required to pump all of the water out of the top of the tank? Assume the same density of water and gravitational constant as in Example \(\PageIndex{4}\).
- Answer
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\(367500\pi\;\) Newton-meters, \(\approx 1150000\) N-m.
Suppose the tank of the previous problem is lying on its side, so that the circular ends are vertical, and that it has the same amount of water as before. How much work is required to pump the water out the top of the tank, which is now 2 meters above the bottom of the tank?
Hint. As this exercise involves a non-trivial integration, here is a possible way to proceed. Measuring height \(h\) positively upwards from \(h=0\) at the bottom so that \(h=1\) is halfway up to the central axis of the tank, proceed as before to find \[W=\sigma g\int_0^1\sqrt{h}(2-h)^{3/2}\,dh.\nonumber\] To evaluate this integral, one way could be to use a combination of partial integrations (twice) and the substitutions \(h=x^2\) and then \(x=\sqrt{2}\cos\theta\), resulting in an expression involving \(\displaystyle\int\sin^6\theta\,d\theta\), which integral can be worked out (amongst other possibilities) as \[\int\sin^6\theta\,d\theta=-\dfrac{\cos\theta}{48}(8\sin^5\theta+10\sin^3\theta+15\sin\theta)+\dfrac{15\theta}{48}+C.\nonumber\]
- Answer
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\(49000\pi+\dfrac{196000}{3}\;\) Newton-meters, \(\approx 220000\) N-m.
A water tank has the shape of the bottom half of a sphere with radius \(r=1\) meter. If the tank is full, how much work is required to pump all the water out of the top of the tank?
- Answer
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\(2450\pi\;\) Newton-meters, \(\approx 7700\) N-m.
A spring has constant \(k=10\) kg/s\(^2\). How much work is done in compressing it from its natural length by \(10\) centimeters?
- Answer
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\(0.05\) Newton-meters.
A force of \(2\) Newtons will compress a spring from \(1\) meter (its natural length) to \(0.8\) meters. How much work is required to stretch the spring from \(1.1\) meters to \(1.5\) meters?
- Answer
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\(1.2\) Newton-meters.
A \(20\) meter long steel cable has density \(2\) kilograms per meter, and is hanging straight down. How much work is required to lift the entire cable to the height of its top end? Continue to assume the same value for the gravitational constant as in Example \(\PageIndex{4}\).
- Answer
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\(3920\) Newton-meters.
The cable in the previous problem has a \(100\) kilogram bucket of concrete attached to its lower end. How much work is required to lift the entire cable and bucket to the height of its top end?
- Answer
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\(23520\) Newton-meters.
Consider again the cable and bucket of the previous problem. How much work is required to lift the bucket \(10\) meters by raising the cable \(10\) meters? (The top half of the cable ends up at the height of the top end of the cable, while the bottom half of the cable is lifted \(10\) meters.)
- Answer
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\(12740\) Newton-meters.
Contributors and Attributions
Integrated by Justin Marshall.