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 $$w$$.

Solution

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 $W=\int_{0}^{D}dW=\int_{0}^{D}w\pi R^{2}(H-x)dx=w\pi R^{2}\left[Hx-\frac{x^{2}}{2}\right]_{0}^{D}=w\pi R^{2}\left(HD-\frac{D^{2}}{2}\right).$