Short and Sweet Calculus

7.6 Work

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 7.16, the work \(W\) done by this force is the product of the force and the distance through which the force acts; that is \[\text{work}=\text{Force}\times\text{displacement\ensuremath{\qquad\text{or}\qquad W=Fd}}\]

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 7.17). 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 \[\boxed{W=\int_{a}^{b}dW=\int_{a}^{b}F(x)dx.}\]

Example 7.8. A cylindrical tank of radius \(R\) and height \(H\) is filled with water to a height \(D\). Find the work required to pump all of the water to the rim of the tank. Denote the weight density of water by \(w\).


Consider a layer of thickness \(dx\) at the height \(x\) above the bottom of the tank as shown in Figure 7.18. The volume of this layer is \(\pi R^{2}dx\), so its weight is \(w\pi R^{2}dx\). The work required to pump this layer to the rim of the tank is \[dW=w\pi R^{2}dx\ (H-x).\] As \(x\) varies between \(0\) and \(D\), the total work done in pumping out all of the water is \[W=\int_{0}^{D}dW=\int_{0}^{D}w\pi R^{2}(H-x)dx=w\pi R^{2}\left[Hx-\frac{x^{2}}{2}\right]_{0}^{D}=w\pi R^{2}\left(HD-\frac{D^{2}}{2}\right).\]

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