9.6 Work - Avidemia

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 is the product as in Figure 1, the work $W$ done by this force is the product of the force and the distance through which the force acts; that is $work = Force \times displacement$ $W = F d$

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 (Figure 2). 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 in this interval, and the small work $d W$ done by this force is $d W = F (x) d x$ This is the element of work. The total work is $W = \int_{a}^{b} d W = \int_{a}^{b} F (x) d x .$

Example 1

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 between their centers

r

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

where

G

is a universal constant called 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}

r = R_{1}

m_{1}

is fixed?

Solution 1

The element of work is

\begin{aligned} d W & = F d r \\ = G \frac{m_{1} m_{2}}{r^{2}} d r \end{aligned}

Therefore, the total work is

\begin{aligned} \int_{R_{0}}^{R_{1}} d W & = \int_{R_{0}}^{R_{1}} G \frac{m_{1} m_{2}}{r^{2}} d r \\ = G m_{1} m_{2} {[- \frac{1}{r}]}_{R_{0}}^{R_{1}} \\ = G m_{1} m_{2} (\frac{1}{R_{0}} - \frac{1}{R_{1}}) . \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 $U = - W = - G \frac{m_{1} m_{2}}{R_{0}} .$
Hooke’s law. The force required to stretch a spring $x$ units beyond its natural length is proportional to $x$ (Figure 3). That is, $f (x) = k x$ where $k$ is a constant called the spring constant. This is called Hooke’s law and holds as long as $x$ is not too large.

Figure 3

Example 2

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 2

First we need to find the spring constant. Hooke’s law states

F = k x

Here

F = 5

N and

x = 2

= 0.02

cm. Therefore,

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

Thus

F = 250 x

The work done in stretching from

x = 0.02

m to

x = 0.08

m is

\begin{aligned} W & = \int_{0.02}^{0.08} 250 x d x \\ = {125 x^{2} |}_{0.02}^{0.08} \\ = 125 ({0.08}^{2} - {0.02}^{2}) \\ = 0.75 J . \end{aligned}

Example 3

A cylindrical tank of radius

R

and height

H

is filled with water to a height

D

(Figure 4). Find the work required to pump all of the water to the rim of the tank. Denote the weight density of water by

w

Figure 4

Solution 3

Consider a layer of thickness

d x

at the height

x

above the bottom of the tank as shown in Figure 5. The volume of this layer is

π R^{2} d x

, so its weight is

w π R^{2} d x

. The work required to pump this layer to the rim of the tank is

d W = w π R^{2} d x (H - x) .

x

varies between

0

and

D

, the total work done in pumping out all of the water is

\begin{aligned} W & = \int_{0}^{D} d W \\ = \int_{0}^{D} w π R^{2} (H - x) d x \\ = w π R^{2} {[H x - \frac{x^{2}}{2}]}_{0}^{D} \\ = w π R^{2} (H D - \frac{D^{2}}{2}) . \end{aligned}

Figure 5

Example 4

An inverted circular cone-shaped tank with height

H

and base radius

R

is filled with water to a height of

D

(Figure 6). Find the work required to pump all of the water to the rim of the tank. Denote the weight density of water by

w

Figure 6

Solution 4

Consider a layer of thickness

d x

at a height

x

above the bottom of the tank as shown in Figure 7(a). This thin layer has the shape of a circular cylinder with radius

r

. From similar triangles (Figure 7(b)), we have

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


(a)	(b)

Figure 8

The volume of this layer is $π r^{2} d x = π {(\frac{R}{H} x)}^{2} d x$ and its weight is $w π (R x / H)^{2} d x$ . The work required to pump this thin layer to the rim of the tank is $d W = w π {(\frac{R}{H} x)}^{2} d x (H - x)$ and because $x$ varies from $0$ to $D$ , the total work is $\begin{aligned} W & = \int_{0}^{D} d W \\ = w π \frac{R^{2}}{H^{2}} \int_{0}^{D} x^{2} (H - x) d x \\ = w π \frac{R^{2}}{H^{2}} {[\frac{H}{3} x^{3} - \frac{1}{4} x^{4}]}_{0}^{D} \\ = w π \frac{R^{2} D^{3}}{H^{2}} (\frac{H}{3} - \frac{D}{4}) . \end{aligned}$