# 2.1: Work

- Page ID
- 89246

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While computing areas and volumes are nice mathematical applications of integration we can also use integration to compute quantities of importance in physics and statistics. One such quantity is work. Work is a way of quantifying the amount of energy that is required to act against a force^{ 1}. In SI^{ 2} metric units the force \(F\) has units newtons (which are kilogram-metres per second squared), \(x\) has units metres and the work \(W\) has units joules (which are newton-metres or kilogram-metres squared per second squared).

The work done by a force \(F(x)\) in moving an object from \(x=a\) to \(x=b\) is

\begin{gather*} W=\int_a^b F(x)\, d{x} \end{gather*}

In particular, if the force is a constant, \(F\text{,}\) independent of \(x\text{,}\) the work is \(F\cdot(b-a)\text{.}\)

Here is some motivation for this definition. Consider a particle of mass \(m\) moving along the \(x\)-axis. Let the position of the particle at time \(t\) be \(x(t)\text{.}\) The particle starts at position \(a\) at time \(\alpha\text{,}\) moves to the right, finishing at position \(b \gt a\) at time \(\beta\text{.}\) While the particle moves, it is subject to a position-dependent force \(F(x)\text{.}\) Then Newton's law of motion ^{3} says ^{4} that force is mass times acceleration

\begin{gather*} m\frac{d^{2}x}{dt^{2}}(t) = F\big(x(t)\big) \end{gather*}

Now consider our definition of work above. It tells us that the work done in moving the particle from \(x=a\) to \(x=b\) is

\begin{align*} W &= \int_a^b F(x) \, d{x} \end{align*}

However, we know the position as a function of time, so we can substitute \(x=x(t)\text{,}\) \(\, d{x}=\dfrac{dx}{dt}\, d{t}\) (using Theorem 1.4.6) and rewrite the above integral:

\[\begin{align*} W = \int_a^b F(x) \, d{x} &= \int_{t=\alpha}^{t=\beta} F(x(t))\dfrac{dx}{dt} \, d{t}\\ \end{align*}\]

Using Newton's second law we can rewrite our integrand:

\begin{align*} &= m \int_\alpha^\beta \frac{d^{2}x}{dt^{2}} \dfrac{dx}{dt} \, d{t}\\ &= m \int_\alpha^\beta \dfrac{dv}{dt} v(t)\, d{t} & \text{since $v(t)=\dfrac{dx}{dt}$}\\ &= m \int_\alpha^\beta \dfrac{d}{dt} \left( \frac{1}{2} v(t)^2 \right) \, d{t} \end{align*}

What happened here? By the chain rule, for any function \(f(t)\text{:}\)

\begin{align*} \dfrac{d}{dt}\left( \frac{1}{2} f(t)^2 \right) &= f(t) f'(t). \end{align*}

In the above computation we have used this fact with \(f(t) = v(t)\text{.}\) Now using the fundamental theorem of calculus (Theorem 1.3.1 part 2), we have

\begin{align*} W&= m \int_\alpha^\beta \dfrac{d}{dt} \left( \frac{1}{2} v(t)^2 \right) \, d{t}\\ &= \frac{1}{2} mv(\beta)^2 - \frac{1}{2}mv(\alpha)^2. \end{align*}

By definition, the function \(\frac{1}{2} mv(t)^2\) is the kinetic energy ^{5} of the particle at time \(t\text{.}\) So the work \(W\) of Definition 2.1.1 is the change in kinetic energy from the time the particle was at \(x=a\) to the time it was at \(x=b\text{.}\)

Imagine that a spring lies along the \(x\)-axis. The left hand end is fixed to a wall, but the right hand end lies freely at \(x=0\text{.}\) So the spring is at its “natural length”.

- Now suppose that we wish to stretch out the spring so that its right hand end is at \(x=L\text{.}\)
- Hooke's Law
^{6}says that when a (linear) spring is stretched (or compressed) by \(x\) units beyond its natural length, it exerts a force of magnitude \(kx\text{,}\) where the constant \(k\) is the spring constant of that spring. - In our case, once we have stretched the spring by \(x\) units to the right, the spring will be trying to pull back the right hand end by applying a force of magnitude \(kx\) directed to the left.
- For us to continue stretching the spring we will have to apply a compensating force of magnitude \(kx\) directed to the right. That is, we have to apply the force \(F(x) = +kx\text{.}\)
- So to stretch a spring by \(L\) units from its natural length we have to supply the work
\begin{align*} W &=\int_0^L k x\, d{x} =\frac{1}{2}kL^2 \end{align*}

A spring has a natural length of \(0.1\)m. If a \(12\)N force is needed to keep it stretched to a length of \(0.12\)m, how much work is required to stretch it from \(0.12\)m to \(0.15\)m?

**Solution:**

In order to answer this question we will need to determine the spring constant and then integrate the appropriate function.

- Our first task is to determine the spring constant \(k\text{.}\) We are told that when the spring is stretched to a length of \(0.12\)m, i.e. to a length of \(0.12-0.1=0.02\)m beyond its natural length, then the spring generates a force of magnitude \(12\)N.
- Hooke's law states that the force exerted by the spring, when it is stretched by \(x\) units, has magnitude \(k x\text{,}\) so
\begin{align*} 12 &= k \cdot 0.02 = k \cdot \frac{2}{100} & \text{thus}\\ k &=600. \end{align*}

- So to stretch the spring
- from a length of \(0.12\)m, i.e. a length of \(x=0.12-0.1=0.02\)m beyond its natural length,
- to a length of \(0.15\)m, i.e. a length of \(x=0.15-0.1=0.05\)m beyond its natural length,

takes work

\begin{align*} W &=\int_{0.02}^{0.05} k x \, d{x} = \left[\frac{1}{2}kx^2\right]_{0.02}^{0.05}\\ &=300\big(0.05^2-0.02^2\big)\\ &=0.63\mathrm{J} \end{align*}

A cylindrical reservoir ^{7} of height \(h\) and radius \(r\) is filled with a fluid of density \(\rho\text{.}\) We would like to know how much work is required to pump all of the fluid out the top of the reservoir.

**Solution:** We are going to tackle this problem by applying the standard integral calculus “slice into small pieces” strategy. This is how we computed areas and volumes — slice the problem into small pieces, work out how much each piece contributes, and then add up the contributions using an integral.

- Start by slicing the reservoir (or rather the fluid inside it) into thin, horizontal, cylindrical pancakes, as in the figure above. We proceed by determining how much work is required to pump out this pancake volume of fluid
^{ 8}. - Each pancake is a squat cylinder with thickness \(\, d{x}\) and circular cross section of radius \(r\) and area \(\pi r^2\text{.}\) Hence it has volume \(\pi r^2 \, d{x}\) and mass \(\rho \times \pi r^2\, d{x}\text{.}\)
- Near the surface of the Earth gravity exerts a downward force of \(mg\) on a body of mass \(m\text{.}\) The constant \(g=9.8\)m/\(\mathrm{sec}^2\) is called the
*standard acceleration due to gravity*^{9}. For us to raise the pancake we have to apply a compensating upward force of \(mg\text{,}\) which, for our pancake, is\begin{align*} F &= g \rho \times \pi r^2\, d{x} \end{align*}

- To remove the pancake at height \(x\) from the reservoir we need to raise it to height \(h\text{.}\) So we have to lift it a distance \(h-x\) using the force \(F=\pi \rho g r^2\, d{x}\text{,}\) which takes work \(\pi\rho g r^2\,(h-x)\, \, d{x}\text{.}\)
- The total work to empty the whole reservoir is
\begin{align*} W&= \int_0^h \pi\,\rho g\,r^2 (h-x)\, d{x} = \pi\,\rho g\,r^2 \int_0^h (h-x)\, d{x}\\ &=\pi\,\rho g\,r^2 \Big[hx -\frac{x^2}{2}\Big]_0^h\\ &=\frac{\pi}{2}\,\rho g\, r^2 h^2 \end{align*}

- If we measure lengths in metres and mass in kilograms, then this quantity has units of Joules. If we instead used feet and pounds
^{ 10}then this would have units of “foot-pounds”. One foot-pound is equal to 1.355817… Joules.

Suppose that you shoot a probe straight up from the surface of the Earth — at what initial speed must the probe move in order to escape Earth's gravity?

**Solution:** We determine this by computing how much work must be done in order to escape Earth's gravity. If we assume that all of this work comes from the probe's initial kinetic energy, then we can establish the minimum initial velocity required.

- The work done by gravity when a mass moves from the surface of the Earth to a height \(h\) above the surface is
\begin{align*} W &= \int_0^h F(x) \, d{x} \end{align*}

where \(F(x)\) is the gravitational force acting on the mass at height \(x\) above the Earth's surface. - The gravitational force
^{ 11}of the Earth acting on a particle of mass \(m\) at a height \(x\) above the surface of the Earth is\begin{gather*} F=-\frac{GMm}{(R+x)^2}, \end{gather*}

where \(G\) is the gravitational constant, \(M\) is the mass of the Earth and \(R\) is the radius of the Earth. Note that \(R+x\) is the distance from the object to the center of the Earth. Additionally, note that this force is negative because gravity acts downward. - So the work done by gravity on the probe, as it travels from the surface of the Earth to a height \(h\text{,}\) is \[\begin{align*} W&=-\int_0^h \frac{GMm}{(R+x)^2}\, d{x}\\ &=-GMm\int_0^h \frac{1}{(R+x)^2}\, d{x}\\ \end{align*}\]
A quick application of the substitution rule with \(u=R+x\) gives

\begin{align*} &=-GMm \int_{u(0)}^{u(h)} \frac{1}{u^2} \, d{u}\\ &= -GMm \left[ -\frac{1}{u} \right]_{u=R}^{u=R+h}\\ &= \frac{GMm}{R+h} - \frac{GMm}{R} \end{align*} - So if the probe completely escapes the Earth and travels all the way to \(h=\infty\text{,}\) gravity does work
\begin{gather*} \lim_{h\rightarrow\infty}\Big[\frac{GMm}{R+h} - \frac{GMm}{R}\Big] =- \frac{GMm}{R} \end{gather*}

The minus sign means that gravity has removed energy \(\frac{GMm}{R}\) from the probe. - To finish the problem we need one more assumption. Let us assume that all of this energy comes from the probe's initial kinetic energy and that the probe is not fitted with any sort of rocket engine. Hence the initial kinetic energy \(\frac{1}{2}mv^2\) (coming from an initial velocity \(v\)) must be at least as large as the work computed above. That is we need
\begin{align*} \frac{1}{2}mv^2 &\ge \frac{GMm}{R} & \text{which rearranges to give}\\ v &\ge \sqrt{\frac{2GM}{R}} \end{align*}

- The right hand side of this inequality, \(\sqrt{\frac{2GM}{R}}\text{,}\) is called the escape velocity.

A \(10\)-metre-long cable of mass \(5\)kg is used to lift a bucket of water, with mass 8kg, out of a well. Find the work done.

**Solution:** Denote by \(y\) the height of the bucket above the top of the water in the well. So the bucket is raised from \(y=0\) to \(y=10\text{.}\) The cable has mass density \(0.5\)kg/m. So when the bucket is at height \(y\text{,}\)

- the cable that remains to be lifted has mass \(0.5(10-y)\) kg and
- the remaining cable and water is subject to a downward gravitational force of magnitude \(\big[0.5(10-y) + 8\big]g=\big[13-\frac{y}{2}\big]g\text{,}\) where \(g=9.8\) m/sec\(^2\text{.}\)

So to raise the bucket from height \(y\) to height \(y+\, d{y}\) we need to apply a compensating upward force of \(\big[13-\frac{y}{2}\big]g\) through distance \(\, d{y}\text{.}\) This takes work \(\big[13-\frac{y}{2}\big]g\, d{y}\text{.}\) So the total work required is

\[ \int_0^{10}\Big[13-\frac{y}{2}\Big]g\, d{y} =g\left[13 y-\frac{y^2}{4}\right]_0^{10} =\big[130-25\big]g =105 g\ \mathrm{J} \nonumber \]

## Exercises

Stage 1

Find the work (in joules) required to lift a 3-gram block of matter a height of 10 centimetres against the force of gravity (with \(g=9.8\) m/sec\(^2\)).

A rock exerts a force of 1 N on the ground where it sits due to gravity. Use \(g=9.8\) m/sec\(^2\text{.}\)

What is the mass of the rock?

How much work (in joules) does it take to lift that rock one metre in the air?

Consider the equation

\[ W = \int_a^b F(x)\,\, d{x} \nonumber \]

where \(x\) is measured in metres and \(F(x)\) is measured in kilogram-metres per second squared (newtons).

For some large \(n\text{,}\) we might approximate

\[ W \approx \sum_{i=1}^n F(x_i)\Delta x \nonumber \]

where \(\Delta x = \frac{b-a}{n}\) and \(x_i\) is some number in the interval \([a+(i-1)\Delta x, a+i\Delta x]\text{.}\) (This is just the general form of a Riemann sum).

- What are the units of \(\Delta x\text{?}\)
- What are the units of \(F(x_i)\text{?}\)
- Using your answers above, what are the units of \(W\text{?}\)

Remark: we already know the units of \(W\) from the text, but the Riemann sum illustrates *why they make sense* arising from this particular integral.

Suppose \(f(x)\) has units \(\dfrac{\mathrm{smoot}}{\mathrm{megaFonzie}}\text{,}\) and \(x\) is measured in barns^{ 12}. What are the units of the quantity \(\int_0^1 f(x)\,\, d{x}\text{?}\)

You want to weigh your luggage before a flight. You don't have a scale or balance, but you do have a heavy-duty spring from your local engineering-supply store. You nail it to your wall, marking where the bottom hangs. You hang a one-litre bag of water (with mass one kilogram) from the spring, and observe that the spring stretches 1 cm. Where on the wall should you mark the bottom of the spring corresponding to a hanging mass of 10kg?

You may assume that the spring obeys Hooke's law.The work done by a force in moving an object from position \(x = 1\) to \(x = b\) is \(W(b) = -b^3+6b^2-9b+4\) for any \(b\) in \([1,3]\text{.}\) At what position \(x\) in \([1,3]\) is the force the strongest?

Stage 2

Questions 9 through 16 offer practice on two broad types of calculations covered in the text: lifting things against gravity, and stretching springs. You may make the same physical assumptions as in the text: that is, springs follow Hooke's law, and the acceleration due to gravity is a constant \(-9.8\) metres per second squared.

For Questions 18 and 19, use the principle (introduced after Definition 2.1.1 and utilized in Example 2.1.5) that the work done on a particle by a force over a distance is equal to the change in kinetic energy of that particle.

A variable force \(F(x) = \frac{a}{\sqrt{x}}\) Newtons moves an object along a straight line when it is a distance of \(x\) meters from the origin. If the work done in moving the object from \(x = 1\) meters to \(x = 16\) meters is \(18\) joules, what is the value of \(a\text{?}\) Don't worry about the units of \(a\text{.}\)

A tube of air is fitted with a plunger that compresses the air as it is pushed in. If the natural length of the tube of air is \(\ell\text{,}\) when the plunger has been pushed \(x\) metres past its natural position, the force exerted by the air is \(\frac{c}{\ell-x}\) N, where \(c\) is a positive constant (depending on the particulars of the tube of air) and \(x \lt \ell\text{.}\)

- What are the units of \(c\text{?}\)
- How much work does it take to push the plunger from 1 metre past its natural position to 1.5 metres past its natural position? (You may assume \(\ell \gt 1.5\text{.}\))

Find the work (in joules) required to stretch a string \(10\) cm beyond equilibrium, if its spring constant is \(k=50\ \mathrm{N}/\mathrm{m}\text{.}\)

A force of \(10\) N (newtons) is required to hold a spring stretched \(5\) cm beyond its natural length. How much work, in joules (J), is done in stretching the spring from its natural length to \(50\) cm beyond its natural length?

A \(5\)-metre-long cable of mass \(8\) kg is used to lift a bucket off the ground. How much work is needed to raise the entire cable to height \(5\) m? Ignore the mass of the bucket and its contents.

A tank 1 metre high has pentagonal cross sections of area 3 m\(^2\) and is filled with water. How much work does it take to pump out all the water?

You may assume the density of water is 1 kg per 1000 cm\(^3\text{.}\)

A sculpture, shaped like a pyramid \(3\)m high sitting on the ground, has been made by stacking smaller and smaller (very thin) iron plates on top of one another. The iron plate at height \(z\) m above ground level is a square whose side length is \((3-z)\) m. All of the iron plates started on the floor of a basement \(2\) m below ground level.

Write down an integral that represents the work, in joules, it took to move all of the iron from its starting position to its present position. Do *not* evaluate the integral. (You can use \(9.8\) m\({}/{}\)s\({}^2\) for the acceleration due to gravity and \(8000\) kg\({}/{}\)m\({}^3\) for the density of iron.)

Suppose a spring extends 5 cm past its natural length when one kilogram is hung from its end. How much work is done to extend the spring from 5 cm past its natural length to 7 cm past its natural length?

Ten kilograms of firewood are hoisted on a rope up a height of 4 metres to a second-floor deck. If the total work done is \(400\) joules, what is the mass of the 4 metres of rope?

You may assume that the rope has the same density all the way along.

A 5 kg weight is attached to the middle of a 10-metre long rope, which dangles out a window. The rope alone has mass 1 kg. How much work does it take to pull the entire rope in through the window, together with the weight?

A box is dragged along the floor. Friction exerts a force in the opposite direction of motion from the box, and that force is equal to \(\mu \times m \times g\text{,}\) where \(\mu\) is a constant, \(m\) is the mass of the box and \(g\) is the acceleration due to gravity. You may assume \(g=9.8\) m/sec\(^2\text{.}\)

- How much work is done dragging a box of mass 10 kg along the floor for three metres if \(\mu=0.4\text{?}\)
- Suppose the box contains a volatile substance that rapidly evaporates. You pull the box at a constant rate of 1 m/sec for three seconds, and the mass of the box at \(t\) seconds (\(0 \leq t \leq 3\)) is \((10-\sqrt{t})\) kilograms. If \(\mu=0.4\text{,}\) how much work is done pulling the box for three seconds?

A ball of mass 1 kg is attached to a spring, and the spring is attached to a table. The ball moves with some initial velocity, and the spring slows it down. At its farthest, the spring stretches 10 cm past its natural length. If the spring constant is 5 N/m, what was the initial velocity of the ball?

You may assume that the ball starts moving with initial velocity \(v_0\text{,}\) and that the only force slowing it down is the spring. You may also assume that the spring started out at its natural length, it follows Hooke's law, and when it is stretched its farthest, the velocity of the ball is 0 m/sec.

A mild-mannered university professor who is definitely not a spy notices that when their car is on the ground, it is 2 cm shorter than when it is on a jack. (That is: when the car is on a jack, its struts are at their natural length; when on the ground, the weight of the car causes the struts to compress 2 cm.) The university professor calculates that if they were to jump a local neighborhood drawbridge, their car would fall to the ground with a speed of 4 m/sec. If the car can sag 20 cm before important parts scrape the ground, and the car has mass 2000 kg unoccupied (2100 kg with the professor inside), can the professor, who is certainly not involved in international intrigue, safely jump the bridge?

Assume the car falls vertically, the struts obey Hooke's law, and the work done by the struts is equal to the change in kinetic energy of the car + professor. Use 9.8 m/sec\(^2\) for the acceleration due to gravity.

Stage 3

A disposable paper cup has the shape of a right circular cone with radius 5 cm and height 15 cm, and is completely filled with water. How much work is done sucking all the water out of the cone with a straw?

You may assume that 1 m\(^3\) of water has mass 1000 kilograms, the acceleration due to gravity is \(-9.8\) m/sec\(^2\text{,}\) and that the water moves as high up as the very top of the cup and no higher.

A spherical tank of radius \(3\) metres is half-full of water. It has a spout of length \(1\) metre sticking up from the top of the tank. Find the work required to pump all of the water in the tank out the spout. The density of water is \(1000\) kilograms per cubic metre. The acceleration due to gravity is \(9.8\) metres per second squared.

A 5-metre cable is pulled out of a deep hole, where it was dangling straight down. The cable has density \(\rho(x) = (10-x)\) kg/m, where \(x\) is the distance from the bottom end of the rope. (So, the bottom of the cable is denser than the top.) How much work is done pulling the cable out of the hole?

A rectangular tank is fitted with a plunger that can raise and lower the water level by decreasing and increasing the length of its base, as in the diagrams below. The tank has base width 1 m (which does not change) and contains 3 m\(^3\) of water.

The force of the water acting on any tiny piece of the plunger is \(PA\text{,}\) where \(P\) is the pressure of the water, and \(\, d{A}\) is the area of the tiny piece. The pressure varies with the depth of the piece (below the surface of the water). Specifically, \(P=cD\text{,}\) where \(D\) is the depth of the tiny piece and \(c\) is a constant, in this case \(c=9800\) N/m\(^3\text{.}\)

- If the length of the base is 3 m, give the force of the water on the entire plunger. (You can do this with an integral: it's the sum of the force on all the tiny pieces of the plunger.)
- If the length of the base is \(x\) m, give the force of the water on the entire plunger.
- Give the work required to move the plunger in so that the base length changes from 3 m to 1 m.

A leaky bucket picks up 5 L of water from a well, but drips out 1 L every ten seconds. If the bucket was hauled up 5 metres at a constant speed of 1 metre every two seconds, how much work was done?

Assume the rope and bucket have negligible mass and one litre of water has 1 kg mass, and use \(9.8\) m/sec\(^2\) for the acceleration due to gravity.

The force of gravity between two objects, one of mass \(m_1\) and another of mass \(m_2\text{,}\) is \(F = G\dfrac{m_1m_2}{r^2}\text{,}\) where \(r\) is the distance between them and \(G\) is the gravitational constant.

How much work is required to separate the earth and the moon far enough apart that the gravitational attraction between them is negligible?

Assume the mass of the earth is \(6 \times 10^{24}\) kg and the mass of the moon is \(7 \times 10^{22}\) kg, and that they are currently \(400\,000\) km away from each other. Also, assume \(G = 6.7\times 10^{-11} \ \frac{\text{m}^3}{\text{kg}\cdot\text{sec}^2}\text{,}\) and the only force acting on the earth and moon is the gravity between them.

True or false: the work done pulling up a dangling cable of length \(\ell\) and mass \(m\) (with uniform density) is the same as the work done lifting up a ball of mass \(m\) a height of \(\ell/2\text{.}\)

A tank one metre high is filled with watery mud that has settled to be denser at the bottom than at the top.

At height \(h\) metres above the bottom of the tank, the cross-section of the tank has the shape of the finite region bounded by the two curves \(y=x^2\) and \(y=2-h-3x^2\text{.}\) At height \(h\) metres above the bottom of the tank, the density of the liquid is \(1000\sqrt{2-h}\) kilograms per cubic metre.

How much work is done to pump all the liquid out of the tank?

You may assume the acceleration due to gravity is \(9.8\) m/sec\(^2\text{.}\)

An hourglass is 0.2 m tall and shaped such that that \(y\) metres above or below its vertical center it has a radius of \(y^2+0.01\) m.

It is exactly half-full of sand, which has mass \(M=\frac{1}{7}\) kilograms.

How much work is done on the sand by quickly flipping the hourglass over?

Assume that the work done is only moving against gravity, with \(g=9.8\) m/sec\(^2\text{,}\) and the sand has uniform density. Also assume that at the instant the hourglass is flipped over, the sand has not yet begun to fall, as in the picture above.

Suppose at position \(x\) a particle experiences a force of \(F(x) = \sqrt{1-x^4}\) N. Approximate the work done moving the particle from \(x=0\) to \(x=1/2\text{,}\) accurate to within \(0.01\) J.

- For example — if your expensive closed-source textbook has fallen on the floor, work quantifies the amount of energy required to lift the object from the floor acting against the force of gravity.
- SI is short for “le système international d'unités” which is French for “the international system of units”. It is the most recent internationally sanctioned version of the metric system, published in 1960. It aims to establish sensible units of measurement (no cubic furlongs per hogshead-Fahrenheit). It defines seven base units — metre (length), kilogram (mass), second (time), kelvin (temperature), ampere (electric current), mole (quantity of substance) and candela (luminous intensity). From these one can then establish derived units — such as metres per second for velocity and speed.
- Specifically, the second of Newton's three law of motion. These were first published in 1687 in his “Philosophiæ Naturalis Principia Mathematica”.
- It actually says something more graceful in Latin - Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur. Or — The alteration of motion is ever proportional to the motive force impressed; and is made in the line in which that force is impressed. It is amazing what you can find on the internet.
- This is not a physics text so we will not be too precise. Roughly speaking, kinetic energy is the energy an object possesses due to it being in motion, as opposed to potential energy, which is the energy of the object due to its position in a force field. Leibniz and Bernoulli determined that kinetic energy is proportional to the square of the velocity, while the modern term “kinetic energy” was first used by Lord Kelvin (back while he was still William Thompson).
- Robert Hooke (1635–1703) was an English contemporary of Isaac Newton (1643–1727). It was in a 1676 letter to Hooke that Newton wrote “If I have seen further it is by standing on the shoulders of Giants.” There is some thought that this was sarcasm and Newton was actually making fun of Hooke, who had a spinal deformity. However at that time Hooke and Newton were still friends. Several years later they did have a somewhat public falling-out over some of Newton's work on optics.
- We could assign units to these measurements — such as metres for the lengths \(h\) and \(r\), and kilograms per cubic metre for the density \(ρ\).
- Potential for a bad “work out how much work out” pun here.
- This quantity is not actually constant — it varies slightly across the surface of earth depending on local density, height above sea-level and centrifugal force from the earth's rotation. It is, for example, slightly higher in Oslo and slightly lower in Singapore. It is actually
*defined*to be \(9.80665\) m/\(\mathrm{sec}^2\) by the International Organisation for Standardization. - It is extremely mysterious to both authors why a person would do science or engineering in imperial units. One of the authors still has nightmares about having had to do so as a student. The other author is thankful that he escaped such tortures.
- Newton published his inverse square law of universal gravitation in his Principia in 1687. His law states that the gravitational force between two masses \(m_1\) and \(m_2\) is \(F = -G \frac{m_1 m_2}{r^2}\) where \(r\) is the distance separating the (centers of the) masses and \(G= 6.674\times 10^{-11} \mathrm{N}\mathrm{m}^2/\mathrm{kg}^2\) is the gravitational constant. Notice that \(r\) measures the separation between the centers of the masses not the distance between the surfaces of the objects. Also, do not confuse \(G\) with \(g\) — standard acceleration due to gravity. The first measurement of \(G\) was performed by Henry Cavendish in 1798 — the interested reader should look up the “Cavendish experiment” for details of this very impressive work.
- For this problem, it doesn't matter what the units measure, but a smoot is a silly measure of length; a megaFonzie is an apocryphal measure of coolness; and a barn is a humorous (but actually used) measure of area. For explanations (and entertainment) see
`https://en.Wikipedia.org/wiki/List_of_humorous_units_of_measurement`

and`https://en.Wikipedia.org/wiki/List_of_unusual_units_of_measurement`

(accessed 27 July 2017).