일

(a) We defined $\Delta x = \frac{b-a}{n}$: that is, the length of one interval, when we chop $[a,b]$ into $n$ of them. If $b$ and $a$ are measured in metres, then $\Delta x$ is measured in metres as well. So, the units of $\Delta x$ are metres.

Put another way, since $a$ and $b$ both describe a quantity in metres, $b-a$ describes a quantity in metres as well. (When we add or subtract quantities of the same units, their sum or difference is given in the same units.) Since $n$ is a unitless quantity (simply a number: not “$n$ kg" or “$n$ m"), $\frac{b-a}{n}$ still describes a quantity in metres. (If I have 6 metres of cloth, and I cut it into 3 pieces, each piece has $\frac{6}{3}=2$ metres–not 2 kilograms, or 2 metres per second.) (b) Since $F(x)$ is measured in kilogram-metres per second squared (newtons), the units of $F(x_i)$ are kilogram-metres per second squared (newtons). (c) $W$ is calculated by adding up summands of the form $F(x_i)\Delta x$. The units of $F(x_i)\Delta x$ are the products of the units of $F(x_i)$ with the units of $\Delta x$. That is, the units of $F(x_i)\Delta x$ are $\left(\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\right)\left(\text{m}\right) = \frac{\text{kg}\cdot\text{m}^2}{\text{sec}^2} = J$. The sum of terms given in joules is itself given in joules, so the units of $W$ are joules.

Hooke's law says that the force required to stretch a spring $x$ units past its natural length is proportional to $x$; that is, there is some constant $k$ associated with the individual spring such that the force required to stretch it $x$ m past its natural length is $kx$.

• Solution 1: Since the force required to stretch the spring is proportional to the amount stretched, and the force acting on the spring is proportional to the mass hanging from it, we conclude the amount the spring stretches is proportional to the mass hung from it. So, if 1 kg stretches it 1 cm, then 10 kg will stretch it 10 cm. We should mark the wall 10 cm below the bottom of the spring as it hangs unloaded.
• Solution 2: We can find $k$ from the test with the bag of water. The force exerted by the bag of water was $(1\text{ kg})(9.8\text{ m/sec}^2) = 9.8$ N$=k(1 \text{ cm})$. So, $$k = \frac{9.8~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}}{0.01\text{ m}} = 980 \frac{\text{kg}}{\text{sec}^2}$$

If we hang 10 kg from the spring, gravity exerts a force of $(10 \text{ kg})(9.8 \text{ m/sec}^2) = 98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}$. This will be matched by the spring with a force of $kx$ newtons, where $k$ is the spring constant and $x$ is the amount stretched.

$$\begin{aligned} kx&=98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\\ \left(980 \frac{\text{kg}}{\text{sec}^2}\right)\left(x\text{ m}\right)&=98~\frac{\text{kg}\cdot\text{m}}{\text{sec}^2}\\ x&=\frac{1}{10}\text{ m} = 10\text{ cm} \end{aligned}$$

So, we should put the mark at 10 cm below the natural length of the spring.

Let's consider sucking up a flat, horizontal layer of water. If the water is $y$ metres above bottom of the cone, then it needs to be raised $0.15-y$ metres. So, if its mass is $m$ kg, then the force of gravity acting on it is $9.8m$ N and the work involved in slurping it to the top of the cone is $9.8m(0.15-y)$ J. So, what we need to find is the mass of a layer of water $y$ metres from the bottom of the cone.

A horizontal cross-section of the cone is a circle. To find its radius, we use similar triangles: $\frac{r}{y} = \frac{0.05}{0.15}$, so $r=\frac{1}{3}y$. Therefore, the area of the cross-section of the cone $y$ metres above its bottom is $\pi\left(\frac{1}{3}y\right)^2 = \frac{\pi}{9}y^2$ m$^2$. If this layer has height $\mathrm{d}y$, then its volume is $\frac{\pi}{9}y^2\,\mathrm{d}y$ m$^3$, and its mass is $1000\frac{\pi}{9}y^2\,\mathrm{d}y$ kg.

Now, we know that the work to suck up the layer of water $y$ metres from the bottom of the cup is $9.8(0.15-y)\left(1000\frac{\pi}{9}y^2\,\mathrm{d}y\right)$ J. So, the work involved in drinking all the water is:

$$\begin{aligned} W&=\int_0^{0.15}9.8(0.15-y)\left(1000\frac{\pi}{9}y^2\right)\,\mathrm{d}y \\ &= \frac{9800\pi}{9}\int_0^{0.15}\big(0.15y^2-y^3\big)\,\mathrm{d}y\\ &= \frac{9800\pi}{9}\left[\frac{0.15}{3}y^3-\frac{1}{4}y^4 \right]_0^{0.15} \\&= \frac{9800\pi}{9}\left[{0.05} (0.15)^3-\frac{1}{4}(0.15)^4 \right]\\ &\approx 0.144\text{ J} \end{aligned}$$

Even drinking water takes work. Life is hard.

(a) The force of the depends on depth, which varies. So, consider a thin rectangle of the plunger at depth $y$, with height $\mathrm{d}y$ and width 1 m (the width of the entire plunger). Let the area of this rectangle be $\mathrm{d}A$.

The area of this rectangle is $1\mathrm{d}y$ m$^2$, so the force of the water acting on it is $F=P\cdot\mathrm{d}A = \underbrace{\left(9800~\tfrac{\text{N}}{\text{m}^3}\right)}_c\underbrace{\left(y\text{ m}\right)}_d \underbrace{\left(\mathrm{d}y \text{ m}^2\right)}_{\mathrm{d}A} = 9800y\,\mathrm{d}y$ N.

The depth at the top of the plunger is $y=0$. To find the depth at the bottom of the plunger, note that the water has a volume of 3 m$^3$, and is in a rectangular container with base 1 m by 3 m. So, its height is 1 m.

The force over the entire plunger, from depth $y=0$ to $y=1$, is $$\int_0^1 9800y\,\mathrm{d}y = \big[4900 y^2\big]_0^1 = 4900 \text{ N}$$ (b) Let's follow our work from part (a), but with the width of the length of the base as $x$ m.

Still, a thin rectangle of plunger has width 1 m and height $\mathrm{d}y$ m, so it has area $\mathrm{d}y$ m$^2$. At depth $y$, it has a force from the water of $9800y\,\mathrm{d}y$ N. This hasn't changed from (a).

Now, let's consider the depth of the water. The volume of water is 3 m$^3$, and it is in a rectangular container with base 1 m by $x$ m. So, its depth is $3/x$ m. Therefore, the force on the entire plunger must be calculated from $y=0$ to $y=3/x$. $$F(x)=\int_0^{3/x} 9800y\,\mathrm{d}y = \big[4900 y^2\big]_0^{3/x} = \frac{9}{x^2}4900 = \frac{44100}{x^2}\text{ N}$$

Let's check that this answer makes sense: $F(3)=4900$ N, which matches our answer from (a). (c) If the force of water acting on the plunger, when the length of the base is $x$ metres, is given by $F(x)$, then we push the plunger with a force of $-F(x)$. Then the work we're looking for is

$$\begin{aligned} W &= \int_3^1 -F(x)\,\mathrm{d}x = \int_1^3 F(x)\,\mathrm{d}x \\ \text{ $F(x)$ is exactly what we found in (b): $F(x) = \frac{44100}{x^2}\text{ N}$.} \\ W&= \int_1^3 F(x)\,\mathrm{d}x = \int_1^3 \frac{44100}{x^2}\,\mathrm{d}x = 44100\left[-\frac{1}{x}\right]_1^3 = 44100\cdot \frac{2}{3} = 29\,400\text{ J} \end{aligned}$$

Let's start by converting from time spent pulling to amount pulled. When $y$ metres of rope have been pulled up, $2y$ seconds have passed, so $\frac{1}{5}y$ litres of water have leaked out of the bucket, leaving $5-\frac{1}{5}y$ litres. (This only makes sense when $\frac{1}{5}y \leq 5$, but we only consider values of $y$ from 0 to 5, so it's not a problem. That is, we're never hauling up an empty bucket that can't leak any more.)

When we've pulled up $y$ metres of rope, the mass in the bucket is $(5-\frac{1}{5}y)$ kg, so the force of gravity acting on it is $9.8(5-\frac{1}{5}y)$ N. Since we pull up 5 metres of rope, the work done is: $$W = \int_0^5 9.8\left(5-\frac{1}{5}y\right) \,\mathrm{d}y = 9.8\left[5y-\frac{1}{10}y^2\right]_0^5 = 9.8\left[25-2.5\right]=220.5\text{ J}$$

According to the formula for gravity between two objects, the earth and moon will gravitationally attract one another no matter how far apart they are, so what we're looking for is the work to separate them infinitely far. That is, we want to calculate $\displaystyle\int_a^\infty F(r)\,\mathrm{d}r$, where $a = 400\,000\,000$ m.

If we take $m_1$ and $m_2$ to be the mass of the earth and moon as given in the question statement, then:

$$\begin{aligned}Gm_1m_2 &= \left(6.7\times 10^{-11} \ \frac{\text{m}^3}{\text{kg}\cdot\text{sec}^2}\right) \left(6 \times 10^{24}\text{ kg}\right)\left(7 \times 10^{22}\text{ kg}\right)\\ &\approx 2.8 \times 10^{37} ~\frac{\text{kg}\cdot\text{m}^3}{\text{sec}^2} \end{aligned}$$

With that out of the way, let's calculate our work.

$$\begin{aligned} W&=\int_a^\infty F(r)\,\mathrm{d}r =\lim_{b \to \infty} \int_a^b G\dfrac{m_1m_2}{r^2}\,\mathrm{d}r \\&= (Gm_1m_2)\lim_{b \to \infty}\left[-\frac{1}{r}\right]_a^b \\ &=(Gm_1m_2)\lim_{b \to \infty} \left(\frac{1}{a}-\frac{1}{b}\right)\\&= \frac{Gm_1m_2}{a}\\ &\approx\frac{2.8 \times 10^{37} ~\frac{\text{kg}\cdot\text{m}^3}{\text{sec}^2}}{4\times 10^8 \text{ m}}\\ &=7 \times 10^{28} \text{ J} \end{aligned}$$

Remark: since the force of gravity between the earth and the moon gets weaker as they are farther apart, it takes less and less work to move them each kilometre. If we move them a finite distance apart, the work involved will always be less than $7\times 10^{28}$ joules, no matter how huge that finite distance is. If we move them a very, very long (but finite) distance apart, the work we did will be quite close to (but still less than) $7\times 10^{28}$ joules.

Since the only work done is against the force of gravity, we only need to know how high the sand was lifted, not how it got there. So, we don't really need to worry about its semicircular path: we can imagine that every grain of sand was lifted from its old position to its new position.

Consider a thin, horizontal layer of sand in the hourglass, $y$ metres below the vertical centre of the hourglass.

• Its final position is $y$ metres above the centre of the hourglass. That is, it was lifted $2y$ metres against the force of gravity.
• The layer is shaped like a circle with radius $y^2+0.01$ and height $\mathrm{d}y$, so its volume is $\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ cubic metres.
• To find the mass of the layer, we need to know the density of the sand. Let the volume of sand in the hourglass be $V$. We are given its mass $M$. Then $\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ cubic metres has a mass of $\frac{M}{V}\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ kilograms.
• So, the force of gravity acting on the layer is $9.8\frac{M}{V}\pi\left(y^2+0.01\right)^2\,\mathrm{d}y$ N, acting vertically downwards.
• To lift the layer to its final position, we apply a compensating force over a distance of $2y$ metres, for a total work of $9.8\frac{M}{V}\pi\left(y^2+0.01\right)^22y\,\mathrm{d}y$ J.
• Since the hourglass has height $0.2$ m, and exactly half of it is filled with sand, the top layer of sand is exactly at the vertical centre of the hourglass, and the bottom layer of sand is 0.1 metres below.

Using $V$ is the volume of sand in the hourglass, and $M$ is its mass (we're given $M =\frac{1}{7}$ kg) then the total work flipping all the sand is:

$$\begin{aligned} W &= \int_0^{0.1} 9.8\frac{M}{V}\pi\left(y^2+0.01\right)^22y\,\mathrm{d}y = \frac{19.6M\pi}{V}\int_0^{0.1}\big(y^5+0.02y^3+0.0001y\big)\,\mathrm{d}y\\ &= \frac{19.6M\pi}{V}\int_0^{1/10}\left(y^5+\frac{2}{10^2}y^3+\frac{1}{10^4}y\right)\,\mathrm{d}y = \frac{19.6M\pi}{V}\left[\frac{1}{6}y^6+\frac{1}{2\cdot10^2}y^4+\frac{1}{2\cdot 10^4}y^2\right]_0^{1/10}\\ &= \frac{19.6M\pi}{V}\left[\left(\frac{1}{6\cdot 10^6}+\frac{1}{2\cdot10^6}+\frac{1}{2\cdot 10^6}\right) - 0\right]\\ &= \frac{19.6M\pi}{V}\left[\frac{7}{6\cdot 10^6}\right]\tag{$*$} \end{aligned}$$

It remains to find $V$: the volume of sand in the hourglass. We know the sand is in the shape of a solid of rotation. Recall from Section 1.6 that we can find the volume of such shapes by slicing them into thin disks.

In the picture above, we've used an axis that matches the way we've been describing our solid: $0$ is the vertical centre of the hourglass, which is where the top of the sand is, and the bottom of the sand is $0.1$ metres from 0.

To find the volume of this solid, we slice it into horizontal disks. The disk that is $x$ metres from the centre of the hourglass has radius $x^2+0.01$ and thickness $\mathrm{d}x$, so it has volume $\pi(x^2+0.01)^2\,\mathrm{d}x$. The volume of the entire solid, i.e. the volume of the sand, is:

$$\begin{aligned}V &= \int_0^{0.1}\pi(x^2+0.01)^2\,\mathrm{d}x = \pi\int_0^{0.1}(x^4+0.02x^2+0.0001)\,\mathrm{d}x\\ &= \pi\left[\frac{1}{5}x^5+\frac{2}{ 3\times10^2}x^3+\frac{1}{10^4}x\right]_0^{0.1} \\&= \pi\left[\left(\frac{1}{5\times 10^5}+\frac{2}{3\times 10^5}+\frac{1}{10^5}\right)-0\right] \\&= \frac{28\pi}{15\times 10^5}= \frac{28\pi}{1.5\times 10^6} \end{aligned}$$

So, the volume of sand in the hourglass is $V=\frac{28\pi}{1.5\times 10^6}$ cubic metres.

Using ($*$), we can find the total work done quickly flipping the hourglass.

$$\begin{aligned} W&=\frac{19.6M\pi}{V}\left[\frac{7}{6\cdot 10^6} \right]\\ &=\frac{19.6 \times \frac{1}{7} \pi}{\frac{28\pi}{1.5\times 10^6}}\left[\frac{7}{6\cdot 10^6} \right] \\&=\frac{19.6 \times 1.5 }{28 \times 6} = \frac{29.4}{168} =\frac{7}{40}= 0.175 \text{ J} \end{aligned}$$

(a) First, we construct a picture of the tank as described. Below we see a sketch of $f(x) = 3\cos \left( \frac{x^3}{4} \right)$ in the setting where the positive $x$-axis points down; note that this function first crosses the $x$-axis at about $1.84527$. We reflect the curve from the first quadrant across the positive $x$-axis to generate a solid of revolution. Cross-sections perpendicular to the $x$-axis have thickness $\triangle x$ and are circular disks. We now suppose that the resulting tank is filled to a depth of $1.5$ feet with water weighing $62.4$ pounds per cubic foot. Since the bottom of the tank is at $1.84527$, being full to a depth of $1.5$ feet means that there is water from $x = 0.34527$ to $x = 1.84527$. To find the work done to pump the water until $0.5$ feet remain in the tank (i.e. to a level where $x = 1.34527$), we have to first find the weight of a typical slice, which requires us to know the volume of such a slice. Observe that for a slice at depth $x$, $$V_{\text{slice}} \approx \pi \left( 3\cos \left( \frac{x^3}{4} \right) \right)^2 \triangle x\,.$$ Since water weighs $62.4$ pounds per cubic foot, the weight (and thus force to be exerted) of such a slice is $$F_{\text{slice}} \approx 62.4 \cdot 9 \cdot \pi \cos^2 \left( \frac{x^3}{4} \right) \triangle x\,.$$ Finally, the work done to pump such a slice to the top of the tank requires the slice to move $x$ units, and thus work for that slice is $$W_{\text{slice}} \approx x \cdot 561.6 \pi \cos^2 \left( \frac{x^3}{4} \right) \triangle x\,.$$ To find the total work accomplished, we use a definite integral to sum the work done on each slice from $x = 0.34527$ (the top of the water when the tank is initially full to $1.84527$ feet) to $x = 1.34527$ (the height of the water after $1$ foot has been emptied). Thus, $$W = \int_{0.34527}^{1.34527} 561.6 \pi \cdot x \cos^2 \left( \frac{x^3}{4} \right) \, dx\,.$$ Evaluating the integral, we find $W \approx 1353.55$ foot-pounds. (b) If instead we envision the tank not as a solid of revolution but as a trough, where the two-dimensional figure above forms one end of a tank that is 10 feet long, we now have a similar problem but with slices that are thin rectangular slabs, rather than thin cylindrical disks. Note that each such slice is $2 f(x)$ wide, $10$ feet long, and $\triangle x$ thick. Thus, for a slice at depth $x$, $$V_{\text{slice}} \approx 2 \left( 3\cos \left( \frac{x^3}{4} \right) \right) \cdot 10 \cdot \triangle x\,.$$ Since water weighs $62.4$ pounds per cubic foot, the weight (and thus force to be exerted) of such a slice is $$F_{\text{slice}} \approx 62.4 \cdot 60 \cos \left( \frac{x^3}{4} \right) \triangle x\,.$$ Finally, the work done to pump such a slice to the top of the tank requires the slice to move $x$ units, and thus work for that slice is $$W_{\text{slice}} \approx x \cdot 3744 \cos \left( \frac{x^3}{4} \right) \triangle x\,.$$ Assuming that the tank is completely full, to lower the tank by $h$ feet, we consider slices from $x = 0$ to $x = h$, and thus using a definite integral to sum the work required for each slice, we find that the total work is $$W = \int_0^h 3744 x \cos \left( \frac{x^3}{4} \right) \, dx\,.$$ (c) For the trough described in (b), we now seek to find the total hydrostatic force on one of the ends of the tank when it is completely full. First, we note that a typical slice at location $x$ will be at a depth of $x$ feet below the water and have area $$A_{\text{slice}} \approx 2 f(x) \triangle x = 2 \left( 3\cos \left( \frac{x^3}{4} \right) \right) \triangle x\,.$$ At any point on this slice, which again lies at depth $x$, the pressure exerted is $$P_{\text{slice}} \approx 62.4 x$$ since pressure is the weight density of water times depth. Finally, since $F = P \cdot A$, the total force water is exerting on this slice is $$F_{\text{slice}} = P_{\text{slice}} \cdot A_{\text{slice}} \approx 62.4 x \cdot 6 \cos \left( \frac{x^3}{4} \right) \triangle x\,.$$ Integrating to sum the total force, we find that $$F = \int_0^{1.84527} 374.4 x \cos \left( \frac{x^3}{4} \right) \, dx \approx 462.637$$ pounds.