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 $\text{work}=\text{Force}\times\text{displacement}$ $W=Fd$

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 $$dx$$, 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.
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 $\frac{r}{R}=\frac{x}{H}\Rightarrow r=\frac{R}{H}x.$

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}