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 \[\text{work}=\text{Force}\times\text{displacement}\]\[W=Fd\]
Figure 1
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.
Figure 2
If the object moves a little bit \(dx\), then we can assume that \(F(x)\) is a constant force in this interval, and the small work \(dW\) done by this force is \[dW=F(x)\ dx\] This is the element of work. The total work is \[ \bbox[#F2F2F2,5px,border:2px solid black]{W=\int_{a}^{b}dW=\int_{a}^{b}F(x)dx.}\]
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 (\(G=6.674\times10^{-11}\text{ m}^{3}\text{kg}^{-1}\text{s}^{-2}\)). How much work is required to move \(m_{2}\) from \(r=R_{0}\) to \(r=R_{1}\) if \(m_{1}\) is fixed?
Solution 1
The element of work is \[\begin{aligned} dW & =Fdr\\ & =G\frac{m_{1}m_{2}}{r^{2}}dr\end{aligned}\] Therefore, the total work is \[\begin{aligned} \int_{R_{0}}^{R_{1}}dW & =\int_{R_{0}}^{R_{1}}G\frac{m_{1}m_{2}}{r^{2}}dr\\ & =Gm_{1}m_{2}\left[-\frac{1}{r}\right]_{R_{0}}^{R_{1}}\\ & =Gm_{1}m_{2}\left(\frac{1}{R_{0}}-\frac{1}{R_{1}}\right).\end{aligned}\]
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 \[U=-W=-G\frac{m_{1}m_{2}}{R_{0}}.\]
Hooke’s law. The force required to stretch a spring \(x\) units beyond its natural length is proportional to \(x\) (Figure 3). That is, \[f(x)=kx\] 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 \[F=kx\] Here \(F=5\) N and \(x=2\) m \(=0.02\) cm. Therefore, \[k=\frac{5}{0.02}=250\ \text{N/m}\] Thus \[F=250x\] The work done in stretching from \(x=0.02\) m to \(x=0.08\) m is \[\begin{aligned} W & =\int_{0.02}^{0.08}250x\ dx\\ & =\left.125x^{2}\right|_{0.02}^{0.08}\\ & =125(0.08^{2}-0.02^{2})\\ & =0.75\text{ J}.\end{aligned}\]
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 \(\gamma \).
Figure 4
Solution 3
Consider a layer of thickness \(dx\) at the height \(x\) above the bottom of the tank as shown in Figure 5. The volume of this layer is \(\pi R^{2}dx\), so its weight is \(\gamma \pi R^{2}dx\). 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 of this layer to the rim of the tank. Therefore, the work done in pumping this later to the top of the tank is \[dW=\underbrace{\gamma\pi R^{2}dx}_{\rm force}\ \underbrace{(H-x)}_{\rm distance}.\] As \(x\) varies between \(0\) and \(D\), the total work done in pumping out all of the water is \[\begin{aligned} W & =\int_{0}^{D}dW\\ & =\int_{0}^{D}\gamma\pi R^{2}(H-x)dx\\ & =\gamma\pi R^{2}\left[Hx-\frac{x^{2}}{2}\right]_{0}^{D}\\ & =\gamma\pi R^{2}\left(HD-\frac{D^{2}}{2}\right).\end{aligned}\]
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 \(\gamma\).
Figure 6
Solution 4
Consider a layer of thickness \(dx\) 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 \(\pi r^{2}dx=\pi\left(\frac{R}{H}x\right)^{2}dx\), so its weight is \(\gamma\pi(Rx/H)^{2}dx\). 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 of this layer to the top of the tank. The work done in pumping this thin layer to the rim of the tank is thus \[dW=\underbrace{\gamma \pi\left(\frac{R}{H}x\right)^{2}dx}_{\rm force}\ \underbrace{(H-x)}_{\rm distance}.\] Because \(x\) increases from \(0\) to \(D\), the total work is \[\begin{aligned} W & =\int_{0}^{D}dW\\ & =\gamma\pi\frac{R^{2}}{H^{2}}\int_{0}^{D}x^{2}(H-x)dx\\ & =\gamma \pi\frac{R^{2}}{H^{2}}\left[\frac{H}{3}x^{3}-\frac{1}{4}x^{4}\right]_{0}^{D}\\ & =\gamma\pi\frac{R^{2}D^{3}}{H^{2}}\left(\frac{H}{3}-\frac{D}{4}\right).\end{aligned}\]
