## 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 \[ \bbox[#F2F2F2,5px,border:2px solid black]{W=\int_{a}^{b}dW=\int_{a}^{b}F(x)dx.}\]

**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