Work

Work is force times distance, but only when the force is constant and acts along the direction of motion. When the force varies, we slice the motion into pieces so small that the force is effectively constant on each piece, add up the tiny contributions $d W = F (x) d x$ , and pass to the integral $W = \int_{a}^{b} F (x) d x$ . The same slicing idea, applied to a tank of liquid instead of a moving object, solves every pumping problem.

Item	Statement
Constant force along a line	$W = F d$
Element of work	$d W = F (x) d x$
Variable force along a line	$W = \int_{a}^{b} F (x) d x$
Unit of work	$1 joule = 1 N \cdot m$ , and $1 N = 1 kg \cdot m / s^{2}$
Hooke's law	$f (x) = k x$ , where $k$ is the spring constant
Stretching a spring from $a$ to $b$	$\int_{a}^{b} k x d x = \frac{k}{2} (b^{2} - a^{2})$
Newton's gravitation	$F = \frac{G m_{1} m_{2}}{r^{2}}$ , $G = 6.674 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2}$
Work against gravity, $R_{0} \to R_{1}$	$G m_{1} m_{2} (\frac{1}{R_{0}} - \frac{1}{R_{1}})$
Gravitational potential energy	$U = - \frac{G m_{1} m_{2}}{R_{0}}$
Pumping a liquid	$W = \int γ A (x) ℓ (x) d x$ , $ℓ (x) =$ lift distance
Weight density of water	$γ \approx 9800 {N/m}^{3}$ (SI); $62.4 {lb/ft}^{3}$ (US)
Cylindrical tank, depth $D$ , pumped to the rim	$γ π R^{2} (H D - \frac{D^{2}}{2})$
Cone (vertex down), depth $D$ , pumped to the rim	$γ π \frac{R^{2} D^{3}}{H^{2}} (\frac{H}{3} - \frac{D}{4})$
Hoisting a rope of linear density $ρ$	$W = \int_{0}^{L} ρ g x d x = \frac{1}{2} ρ g L^{2}$
Compressing a gas, $p V = C$	$W = \int_{V_{1}}^{V_{2}} p d V = C \ln \frac{V_{2}}{V_{1}}$

Work Done by a Constant Force Along a Line

Suppose $F$ is a constant force. If $F$ acts on an object that moves in a straight line in the direction of the force, as in the figure below, the work $W$ done by this force is the product of the force and the distance through which the force acts; that is,

work = force \times displacement

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{W=Fd.}

A constant force F acting on an object that moves in a straight line in the direction of the force through a displacement d. — A constant force $F$ acting through a displacement $d$ in the direction of the force does work $W = F d$ .

If $F$ is measured in newtons ( $N = kg \cdot m / s^{2}$ ) and displacement in meters, then the unit of $W$ is a newton-meter ( $N \cdot m$ ), called a joule (J).

Work Done by a Variable Force Along a Line

Suppose $F$ is a variable force, and it acts in a given direction on an object moving in this direction, which we take to be the $x$ -axis. The work done by $F (x)$ as its point of application moves from $x = a$ to $x = b$ can be computed by integration.

If the object moves a little bit $d x$ , then we can assume that $F (x)$ is a constant force on this tiny interval, and the small work $d W$ done by the force is

d W = F (x) d x .

This is the element of work. Geometrically it is the area of a thin strip of width $d x$ and height $F (x)$ under the graph of the force, as in the figure below. The total work is the sum of all such strips:

\bbox[8px, #E6F0FA, border: 3px solid #0066CC]{W=\int_{a}^{b}dW=\int_{a}^{b}F(x)\,dx.}

A variable force F of x acting along the x-axis on an object whose point of application moves from x = a to x = b. — The work done by a variable force $F (x)$ as its point of application moves from $x = a$ to $x = b$ .

Newton's law of gravitation states that every two particles of masses $m_{1}$ and $m_{2}$ attract each other with a force $F$ that is directly proportional to the product of their masses and inversely proportional to the square of the distance $r$ between their centers,

F = G \frac{m_{1} m_{2}}{r^{2}},

where $G$ is a universal constant called the gravitational constant ( $G = 6.674 \times 10^{- 11} m^{3} {kg}^{- 1} s^{- 2}$ ). How much work is required to move $m_{2}$ from $r = R_{0}$ to $r = R_{1}$ if $m_{1}$ is fixed?

Solution

The element of work is

\begin{aligned} dW &= F\,dr\\ &= G\frac{m_{1}m_{2}}{r^{2}}\,dr. \end{aligned}

Therefore, the total work is

\begin{aligned} W=\int_{R_{0}}^{R_{1}}dW &= \int_{R_{0}}^{R_{1}}G\frac{m_{1}m_{2}}{r^{2}}\,dr\\ &= Gm_{1}m_{2}\left[-\frac{1}{r}\right]_{R_{0}}^{R_{1}}\\ &= Gm_{1}m_{2}\left(\frac{1}{R_{0}}-\frac{1}{R_{1}}\right). \end{aligned}

Gravitational potential energy. The work required to completely separate two particles is the opposite of the gravitational potential energy $U$ . So if in the above example we let $R_{1} \to \infty$ , then $1 / R_{1} \to 0$ and $U = - W = - G \frac{m_{1} m_{2}}{R_{0}} .$
Hooke's law. The force required to stretch a spring $x$ units beyond its natural length is proportional to $x$ , as in the figure below. That is, $f (x) = k x,$
where $k$ is a constant called the spring constant. This is called Hooke's law, and it holds as long as $x$ is not too large.

Hooke's law: the force needed to hold a spring stretched $x$ units beyond its natural length is $f (x) = k x$ .

The natural length of a spring is 10 cm. If we apply a force of 5 N, the length of the spring increases to 12 cm. How much work is done in stretching the spring from 12 cm to 18 cm?

Solution

First we need to find the spring constant. Hooke's law states $F = k x$ , where $x$ is measured from the natural length. Here $F = 5$ N and $x = 2 cm = 0.02$ m. Therefore,

k = \frac{5}{0.02} = 250 N/m,

so $F = 250 x$ . A length of 12 cm means $x = 0.02$ m and a length of 18 cm means $x = 0.08$ m, so the work done in stretching from $x = 0.02$ m to $x = 0.08$ m is

\begin{aligned} W &= \int_{0.02}^{0.08}250x\,dx\\ &= \left.125x^{2}\right|_{0.02}^{0.08}\\ &= 125\left(0.08^{2}-0.02^{2}\right)\\ &= 125(0.0064-0.0004)\\ &= 125(0.006)=0.75\ \text{J}. \end{aligned}

Note that $x$ must be measured from the natural length, not from the floor or from the end of the stretched spring.

Pumping a Liquid Out of a Tank

Every pumping problem in this section is the same problem. The liquid at the bottom of the tank has to travel farther than the liquid at the top, so the lift distance is not a single number, and we cannot simply write $W = F d$ . Instead we slice the liquid into thin horizontal layers, each of which moves essentially the same distance, and integrate.

The recipe.

Draw a vertical $x$ -axis. Unless the problem suggests otherwise, put $x = 0$ at the bottom of the tank with $x$ increasing upward.
Slice the liquid into horizontal layers of thickness $d x$ at height $x$ .
Find the cross-sectional area $A (x)$ of the tank at height $x$ . The volume of the layer is $d V = A (x) d x$ .
Multiply by the weight density $γ$ to get the weight of the layer, which is exactly the force needed to lift it: $d F = γ A (x) d x$ .
Find the lift distance $ℓ (x)$ , the vertical distance from the layer at height $x$ to the outlet. If the liquid leaves at height $h$ , then $ℓ (x) = h - x$ .
Multiply force by distance to get the element of work, $d W = γ A (x) ℓ (x) d x$ , and integrate over the liquid that is actually present, not over the whole tank:

W = \int_{bottom of liquid}^{top of liquid} γ A (x) ℓ (x) d x .

The weight density $γ$ is weight per unit volume, not mass per unit volume. In SI units, water has mass density $1000 kg / m^{3}$ , so its weight density is $γ = ρ g = 1000 \times 9.8 = 9800 N / m^{3} .$
In US customary units the weight density of water is already a weight, $γ = 62.4 lb / {ft}^{3}$ , and no factor of $g$ is inserted.

Two traps to avoid. First, the limits of integration describe the liquid, while the outlet height in $ℓ (x)$ describes the tank (or the spout above it); these are different numbers whenever the tank is not full. Second, $A (x)$ is the area of the horizontal cross section, so for a cone or a sphere it changes with $x$ and must be found by similar triangles or by the Pythagorean theorem.

A cylindrical tank of radius $R$ and height $H$ is filled with water to a height $D$ . How much work is done in pumping all of the water to the rim of the tank? Denote the weight density of water by $γ$ .

A vertical cylindrical tank of radius R and height H holding water to a depth D, with empty space between the water surface and the rim. — A cylindrical tank of radius $R$ and height $H$ , filled with water to a height $D$ .

Solution

Consider a layer of thickness $d x$ at the height $x$ above the bottom of the tank, as shown in the figure. The cross section of a cylinder is the same disk at every height, so the volume of this layer is $π R^{2} d x$ , and its weight is $γ π R^{2} d x$ . The force required to lift this layer is equal to its weight. The distance through which this force must act is $H - x$ , which is the distance from the layer to the rim of the tank. Therefore, the work required to pump this layer out is

d W = \underset{force}{\underset{⏟}{γ π R^{2} d x}} \underset{distance to rim}{\underset{⏟}{(H - x)}} .

The cylindrical tank with a thin horizontal layer of water of thickness dx at height x above the bottom, and the distance H minus x from that layer up to the rim of the tank. — The layer of thickness $d x$ at height $x$ must be lifted a distance $H - x$ to reach the rim.

As $x$ varies between $0$ and $D$ , the total work done in pumping out all of the water is

\begin{aligned} W &= \int_{0}^{D}dW\\ &= \int_{0}^{D}\gamma\pi R^{2}(H-x)\,dx\\ &= \gamma\pi R^{2}\left[Hx-\frac{x^{2}}{2}\right]_{0}^{D}\\ &= \gamma\pi R^{2}\left(HD-\frac{D^{2}}{2}\right). \end{aligned}

As a check, if the tank is full then $D = H$ and $W = γ π R^{2} H^{2} / 2$ . That is the total weight $γ π R^{2} H$ times $H / 2$ , which says that the whole body of water behaves as though it were concentrated at its midpoint, exactly as expected.

An inverted circular cone-shaped tank with height $H$ and base radius $R$ is filled with water to a height of $D$ . Find the work required to pump all of the water to the rim of the tank. Denote the weight density of water by $γ$ .

An inverted cone-shaped tank with its vertex at the bottom, of height H and base radius R, holding water to a depth D. — An inverted cone-shaped tank of height $H$ and base radius $R$ , filled with water to a height $D$ .

Solution

Consider a layer of thickness $d x$ at a height $x$ above the bottom (that is, above the vertex) of the tank, as shown in the figure below.

The inverted cone-shaped tank with a thin horizontal layer of water of thickness dx at height x above the vertex, and the distance H minus x from that layer up to the rim. — The layer of thickness $d x$ at height $x$ is a thin disk of radius $r$ , and it must be lifted a distance $H - x$ .

This thin layer has the shape of a circular cylinder with radius $r$ . From similar triangles (see the figure below),

\frac{r}{R} = \frac{x}{H} \Rightarrow r = \frac{R}{H} x .

A vertical cross section of the cone showing the small triangle of height x and base r sitting inside the large triangle of height H and base R, which gives the ratio r over R equals x over H. — Similar triangles give $\frac{r}{R} = \frac{x}{H}$ .

The volume of this layer is $π r^{2} d x = π {(\frac{R}{H} x)}^{2} d x$ , so its weight is $γ π {(\frac{R}{H} x)}^{2} d x$ . The force required to lift this layer is equal to its weight. The distance through which this force must act is $H - x$ , which is the distance from this layer to the rim of the tank. The work done in lifting this thin layer to the rim is thus

d W = \underset{force}{\underset{⏟}{γ π {(\frac{R}{H} x)}^{2} d x}} \underset{distance}{\underset{⏟}{(H - x)}} .

As $x$ increases from $0$ to $D$ , the total work is

\begin{aligned} W &= \int_{0}^{D}dW\\ &= \gamma\pi\frac{R^{2}}{H^{2}}\int_{0}^{D}x^{2}(H-x)\,dx\\ &= \gamma\pi\frac{R^{2}}{H^{2}}\left[\frac{H}{3}x^{3}-\frac{1}{4}x^{4}\right]_{0}^{D}\\ &= \gamma\pi\frac{R^{2}D^{3}}{H^{2}}\left(\frac{H}{3}-\frac{D}{4}\right). \end{aligned}

Notice that the only thing that changed from the cylinder is the cross-sectional area: $π R^{2}$ became $π (R x / H)^{2}$ . Everything else in the recipe is identical.

Exercises

Throughout, take $g = 9.8 m / s^{2}$ and the weight density of water to be $γ = 9800 N / m^{3}$ .

(a) A crate is pushed $12$ m across a level floor by a constant horizontal force of $250$ N. Find the work done.

(b) A particle moves along the $x$ -axis from $x = 0$ to $x = 4$ m under a force $F (x) = 3 x^{2} + 2 x$ newtons directed along the axis. Find the work done.

Solution

(a) The force is constant and acts along the direction of motion, so

W = F d = 250 \times 12 = 3000 J .

(b) The force varies, so we integrate the element of work $d W = F (x) d x$ :

\begin{aligned} W &= \int_{0}^{4}\left(3x^{2}+2x\right)dx\\ &= \left[x^{3}+x^{2}\right]_{0}^{4}\\ &= 64+16=80\ \text{J}. \end{aligned}

A spring has natural length $20$ cm. A force of $40$ N holds it stretched to a length of $25$ cm. Find the spring constant $k$ , and then find the work done in stretching the spring from its natural length to a length of $30$ cm.

Solution

Measure $x$ from the natural length. A length of $25$ cm corresponds to $x = 0.05$ m, and Hooke's law $F = k x$ gives

k = \frac{40}{0.05} = 800 N/m .

A length of $30$ cm corresponds to $x = 0.10$ m, so

\begin{aligned} W &= \int_{0}^{0.10}800x\,dx\\ &= \left.400x^{2}\right|_{0}^{0.10}\\ &= 400(0.01)=4\ \text{J}. \end{aligned}

It takes $6$ J of work to stretch a certain spring from its natural length to $0.2$ m beyond its natural length. How much work is required to stretch it from $0.2$ m beyond its natural length to $0.4$ m beyond?

Solution

First recover $k$ from the given work. Stretching from $0$ to $0.2$ m requires

\int_{0}^{0.2} k x d x = \frac{k}{2} (0.2)^{2} = 0.02 k,

and this equals $6$ J, so $k = 300 N/m$ . Then

\begin{aligned} W &= \int_{0.2}^{0.4}300x\,dx\\ &= \left.150x^{2}\right|_{0.2}^{0.4}\\ &= 150(0.16-0.04)\\ &= 150(0.12)=18\ \text{J}. \end{aligned}

The second stretch of the same length costs three times as much work as the first, because the force is larger throughout.

A uniform rope of length $20$ m and linear mass density $0.5 kg / m$ hangs over the edge of a tall building. How much work is required to pull the entire rope to the top?

Solution

Let $x$ be the distance measured downward from the top, so the rope occupies $0 \leq x \leq 20$ . A piece of rope of length $d x$ located at depth $x$ has mass $0.5 d x$ kg and therefore weight

d F = (0.5) (9.8) d x = 4.9 d x N .

That piece must be raised a distance $x$ , so the element of work is $d W = 4.9 x d x$ and

\begin{aligned} W &= \int_{0}^{20}4.9x\,dx\\ &= \left.2.45x^{2}\right|_{0}^{20}\\ &= 2.45(400)=980\ \text{J}. \end{aligned}

Equivalently, the whole rope weighs $20 (4.9) = 98$ N and its center is $10$ m below the top, giving $98 \times 10 = 980$ J.

A bucket of mass $2$ kg containing $20$ kg of water is hoisted $15$ m up a well at a constant rate by a rope of negligible mass. Water leaks out at a constant rate, and exactly $5$ kg has escaped by the time the bucket reaches the top. How much work is done?

Solution

Let $x$ be the height in meters above the starting point, $0 \leq x \leq 15$ . The water leaks at a constant rate, losing $5$ kg over $15$ m, that is $\frac{1}{3}$ kg per meter. So at height $x$ the total mass being lifted is

m (x) = \underset{bucket}{\underset{⏟}{2}} + \underset{water}{\underset{⏟}{(20 - \frac{x}{3})}} = 22 - \frac{x}{3} kg,

and the force required is $F (x) = 9.8 (22 - \frac{x}{3})$ newtons. Therefore

\begin{aligned} W &= \int_{0}^{15}9.8\left(22-\frac{x}{3}\right)dx\\ &= 9.8\left[22x-\frac{x^{2}}{6}\right]_{0}^{15}\\ &= 9.8\left(330-\frac{225}{6}\right)\\ &= 9.8(330-37.5)\\ &= 9.8(292.5)=2866.5\ \text{J}. \end{aligned}

Note that the leaked water is not lifted the full $15$ m, which is why the answer is less than $9.8 (22) (15) = 3234$ J.

A rectangular trough is $4$ m long, $2$ m wide, and $1.5$ m deep. It is full of water. Find the work required to pump all the water over the top edge.

Solution

Measure $x$ upward from the bottom of the trough. Every horizontal cross section is the same $4 \times 2$ rectangle, so

A (x) = 8 m^{2},

and a layer of thickness $d x$ weighs $9800 (8) d x = 78400 d x$ newtons. The top edge is at height $1.5$ , so the lift distance is $ℓ (x) = 1.5 - x$ . The water occupies $0 \leq x \leq 1.5$ , so

\begin{aligned} W &= \int_{0}^{1.5}78400(1.5-x)\,dx\\ &= 78400\left[1.5x-\frac{x^{2}}{2}\right]_{0}^{1.5}\\ &= 78400(2.25-1.125)\\ &= 78400(1.125)=88200\ \text{J}. \end{aligned}

A vertical cylindrical tank of radius $2$ m and height $6$ m is full of water. Find the work required to pump all the water to the rim.

Solution

Take $x$ upward from the bottom. The cross section is a disk of radius $2$ , so $A (x) = π (2)^{2} = 4 π m^{2}$ at every height, and a layer of thickness $d x$ weighs $9800 (4 π) d x$ . The rim is at height $6$ , so the lift distance is $6 - x$ , and the water fills $0 \leq x \leq 6$ :

\begin{aligned} W &= \int_{0}^{6}9800(4\pi)(6-x)\,dx\\ &= 39200\pi\left[6x-\frac{x^{2}}{2}\right]_{0}^{6}\\ &= 39200\pi(36-18)\\ &= 705600\pi\approx2.22\times10^{6}\ \text{J}. \end{aligned}

This agrees with the general formula $γ π R^{2} (H D - \frac{D^{2}}{2})$ derived in the text, with $R = 2$ and $H = D = 6$ .

A tank in the shape of an inverted right circular cone (vertex at the bottom) has height $5$ m and top radius $2$ m. It contains water to a depth of $3$ m. Find the work required to pump all the water to the rim.

Solution

Measure $x$ upward from the vertex. By similar triangles the radius of the cross section at height $x$ satisfies

\frac{r}{2} = \frac{x}{5} \Rightarrow r = \frac{2 x}{5},

so $A (x) = π r^{2} = \frac{4 π x^{2}}{25}$ . The rim is at height $5$ , so the lift distance is $5 - x$ , and the water occupies $0 \leq x \leq 3$ :

\begin{aligned} W &= \int_{0}^{3}9800\cdot\frac{4\pi x^{2}}{25}(5-x)\,dx\\ &= 1568\pi\int_{0}^{3}\left(5x^{2}-x^{3}\right)dx\\ &= 1568\pi\left[\frac{5x^{3}}{3}-\frac{x^{4}}{4}\right]_{0}^{3}\\ &= 1568\pi\left(45-20.25\right)\\ &= 1568\pi(24.75)=38808\pi\approx1.22\times10^{5}\ \text{J}. \end{aligned}

Here $9800 \cdot \frac{4 π}{25} = \frac{39200 π}{25} = 1568 π$ . The answer also follows from the general formula $γ π \frac{R^{2} D^{3}}{H^{2}} (\frac{H}{3} - \frac{D}{4})$ with $R = 2$ , $H = 5$ , $D = 3$ .

A tank has the shape of a right circular cone standing on its base, with base radius $3$ m at the bottom, apex at the top, and height $4$ m. The tank is full of water. Find the work required to pump all the water out through an opening at the apex.

Solution

Measure $x$ upward from the base. Now the cross sections shrink as $x$ increases: the radius at height $x$ is $0$ at $x = 4$ and $3$ at $x = 0$ , so by similar triangles

\frac{r}{3} = \frac{4 - x}{4} \Rightarrow r = \frac{3 (4 - x)}{4},

and $A (x) = π r^{2} = \frac{9 π (4 - x)^{2}}{16}$ . The outlet is at the apex, height $4$ , so the lift distance is $4 - x$ , and

\begin{aligned} W &= \int_{0}^{4}9800\cdot\frac{9\pi(4-x)^{2}}{16}(4-x)\,dx\\ &= \frac{88200\pi}{16}\int_{0}^{4}(4-x)^{3}\,dx\\ &= 5512.5\pi\left[-\frac{(4-x)^{4}}{4}\right]_{0}^{4}\\ &= 5512.5\pi\left(0+\frac{256}{4}\right)\\ &= 5512.5\pi(64)=352800\pi\approx1.11\times10^{6}\ \text{J}. \end{aligned}

Compare with the previous exercise: the geometry factor and the lift distance now involve the same quantity $4 - x$ , which is what makes the integrand a perfect cube.

A tank has the shape of a hemispherical bowl of radius $3$ m with its flat circular face on top. It is full of water. Find the work required to pump all the water to the level of the top of the tank.

Solution

Measure $x$ upward from the lowest point of the bowl, so the water occupies $0 \leq x \leq 3$ and the rim is at $x = 3$ . The center of the sphere is at height $3$ , so a horizontal cross section at height $x$ is a circle whose radius $r$ satisfies, by the Pythagorean theorem,

r^{2} = 3^{2} - (3 - x)^{2} = 6 x - x^{2} .

Hence $A (x) = π (6 x - x^{2})$ and the lift distance is $3 - x$ , so

\begin{aligned} W &= \int_{0}^{3}9800\pi\left(6x-x^{2}\right)(3-x)\,dx\\ &= 9800\pi\int_{0}^{3}\left(18x-6x^{2}-3x^{2}+x^{3}\right)dx\\ &= 9800\pi\int_{0}^{3}\left(18x-9x^{2}+x^{3}\right)dx\\ &= 9800\pi\left[9x^{2}-3x^{3}+\frac{x^{4}}{4}\right]_{0}^{3}\\ &= 9800\pi\left(81-81+\frac{81}{4}\right)\\ &= 9800\pi(20.25)=198450\pi\approx6.23\times10^{5}\ \text{J}. \end{aligned}

A vertical cylindrical tank of radius $1$ m and height $4$ m contains water to a depth of $3$ m. The water must be pumped out through a spout whose opening is $1.5$ m above the top of the tank. Find the work required.

Solution

Measure $x$ upward from the bottom of the tank. The spout opening is at height $4 + 1.5 = 5.5$ , so the lift distance for the layer at height $x$ is

ℓ (x) = 5.5 - x .

The cross section is a disk of radius $1$ , so $A (x) = π$ . The water is only $3$ m deep, so the limits are $0$ to $3$ , not $0$ to $4$ :

\begin{aligned} W &= \int_{0}^{3}9800\pi(5.5-x)\,dx\\ &= 9800\pi\left[5.5x-\frac{x^{2}}{2}\right]_{0}^{3}\\ &= 9800\pi(16.5-4.5)\\ &= 9800\pi(12)=117600\pi\approx3.69\times10^{5}\ \text{J}. \end{aligned}

The two numbers that are easy to confuse here are the outlet height, $5.5$ , which enters $ℓ (x)$ , and the water depth, $3$ , which sets the upper limit of integration.

A gas is confined in a cylinder by a movable piston. When the piston moves, the work done by the gas is $W = \int_{V_{1}}^{V_{2}} p d V$ , where $p$ is the pressure and $V$ the volume. Suppose the gas is compressed at constant temperature, so that $p V = C$ for a constant $C$ . If the gas starts at a pressure of $200$ kPa occupying $0.05 m^{3}$ and is compressed to $0.01 m^{3}$ , how much work is done on the gas?

Solution

First find the constant. With $p_{1} = 200 kPa = 2 \times 10^{5} Pa$ and $V_{1} = 0.05 m^{3}$ ,

C = p_{1} V_{1} = (2 \times 10^{5}) (0.05) = 10^{4} N \cdot m .

Since $p V = C$ , we have $p = C / V$ , and the work done by the gas as the volume changes from $V_{1}$ to $V_{2}$ is

\begin{aligned} W_{\text{by}} &= \int_{V_{1}}^{V_{2}}\frac{C}{V}\,dV\\ &= C\left[\ln V\right]_{V_{1}}^{V_{2}}\\ &= C\ln\frac{V_{2}}{V_{1}}. \end{aligned}

With $V_{2} = 0.01$ and $V_{1} = 0.05$ ,

W_{by} = 10^{4} \ln \frac{0.01}{0.05} = 10^{4} \ln \frac{1}{5} = - 10^{4} \ln 5 \approx - 1.61 \times 10^{4} J .

The sign is negative because the gas is compressed rather than allowed to expand. The work done on the gas is the opposite,

W_{on} = 10^{4} \ln 5 \approx 16094 J \approx 1.61 \times 10^{4} J .

Notice that the element of work here is $d W = p d V$ rather than $F d x$ ; if $A$ is the piston area, then $F = p A$ and $d V = A d x$ , so $F d x = p A d x = p d V$ . It is the same formula written in the variable that is convenient.

Work Done by a Constant Force Along a Line

Work Done by a Variable Force Along a Line

Pumping a Liquid Out of a Tank

Exercises

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