1.5: Work
- Page ID
- 163255
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Online Lecture: Force - Definition and Concepts
The Metric and U.S. Customary Systems
| Distance \( d \) |
Mass \( m \) |
Gravitational Constant (at sea level) \( g \) |
Force \( F = m g \) |
Work \( W = F d \) |
|
|---|---|---|---|---|---|
| English System | foot (ft) | slug | \( \approx 32.2 \, \left( \frac{\text{ft}}{\text{s}^2} \right) \) | pound \( \left( \frac{\text{slug-ft}}{\text{s}^2} \right) \) | foot-pound \( \left( \text{ft} \cdot \text{lbs} \right) \) |
| Metric System | meter (m) | kilogram (kg) | \( \approx 9.8 \, \left( \frac{\text{m}}{\text{s}^2} \right) \) | Newton \( \left( \frac{\text{kg m}}{\text{s}^2} \equiv \text{N} \right) \) | joule \( \left( \frac{\text{kg m}}{\text{s}^2} \cdot \text{m} = \text{N m} \right) \) |
Online Lecture: Work - Definitions and Concepts
Work Done by a Variable Force
If a variable force \(F(x)\) moves an object in a positive direction along the \(x\)-axis from point \(a\) to point \(b\), then the work done on the object is\[W=\int ^b_aF(x)\,dx. \nonumber \]
- Proof
When a particle is located a distance \(x\) meters from the origin, a force of \(\cos\left( \frac{\pi x}{3} \right)\) newtons acts on it. How much work is done in moving the particle from \(x=1\) to \(x=2\)? Interpret your answer by considering the work done from \(x=1\) to \(x=1.5\), and from \(x=1.5\) to \(x=2\).
Spring-Mass Problems and Hooke's Law
The force required to maintain a spring stretched \(x\) units beyond its natural length is proportional to \(x\):\[f(x)=kx,\nonumber \]where \(k \geq 0\) is called the spring constant.
This holds providing \( x \) is not too large.
A spring has a natural length of 40 cm. A 60-newton force is required to keep the spring compressed 10 cm.
- How much work is done during this compression?
- How much work is required to compress the spring to a length of 25 cm?
Work - Lifting Solids
A bucket with a mass of 140 kg when filled with water is lifted at a constant rate, by a mechanical winch, from the bottom of a well that is 60 m deep.
- Compute the work required to lift the bucket from the bottom of the well to the top.
- Oops, we forgot to tell you! The chain that is being used to lift the bucket has a mass of 1.1 kg per meter. Fnd the correct amount of work required to lift the bucket from the bottom of the well all the way to the top.
- Oh no! Where is our head at today? We completely forgot to mention that the bucket has a hole in the bottom. It only has a mass of 60 kg when it reaches the top due to the water leaking out at a constant rate. Now compute the actual work required to lift the bucket from the bottom of the well to the top. (Recall that the bucket is being lifted at a constant rate, as well.)
Work - Lifting Fluids
The volumetric density (referred to as density) of a three-dimensional object is its mass per unit volume. In some disciplines, this is also known as the object's specific mass. If we let \( \rho \), \( m \), and \( V \) be the density, mass, and volume, respectively, then\[ \rho = \dfrac{m}{V} \implies m = \rho V. \nonumber \]
What most people refer to as "density" in the U.S. Customary System is the weight density (also referred to as the specific weight) of the object. That is, since density is mass per unit volume, the units for density in the U.S. Customary System are technically slugs per cubic foot; however, you would search far and wide before you found anyone stating that the density of water is approximately \( 1.9 \text{ slugs}/\text{ft}^3\). Most people, even in scientific circles, cite the density of water (at \( 32^{ \circ } \, F \) as approximately \( 62.4 \text{ lbs}/\text{ft}^3 \).
This "loose" definition of density is scientifically inaccurate as the pound is a force and, therefore, the quantity being used is force per unit volume - not mass per unit volume. To avoid confusion, I use "weight density" when referring to force per unit volume. The Greek letter \( \gamma \) (lowercase "gamma") is traditionally used when referring to the weight density of an object, so I will use that convention in this text as well.
| Distance \( d \) |
Mass \( m \) |
Gravitational Constant (at sea level) \( g \) |
Force \( F = mg \) |
Density \( \rho = \frac{\text{m}}{\text{d}^3} \) |
Weight Density \( \gamma = \rho\, g = \frac{\text{F}}{\text{d}^3} \) |
|
|---|---|---|---|---|---|---|
| English System | foot (ft) | slug | \( \approx 32.2 \, \left( \frac{\text{ft}}{\text{s}^2} \right) \) | pound \( \left( \frac{\text{slug-ft}}{\text{s}^2} \right) \) | \( \left( \frac{\text{slugs}}{\text{ft}^3} \right) \) | \( \left( \frac{\text{lbs}}{\text{ft}^3} \right) \) |
| Metric System | meter (m) | kilogram (kg) | \( \approx 9.8 \, \left( \frac{\text{m}}{\text{s}^2} \right) \) | Newton \( \left( \frac{\text{kg m}}{\text{s}^2} \equiv \text{N} \right) \) | \( \left( \frac{\text{kg}}{\text{m}^3} \right) \) | \( \left( \frac{\text{N}}{{m}^3} \right) \) |
Find the work required to pump all the water out of the top of a filled tank that is an upright cylinder of radius 3 feet and height 10 feet and has a hemispherical cap at the top (also filled with water). The weight density of water is 62.4 lbs per cubic foot.
Let's do the last one from the homework set.