9.3.4 Variable Force

We’ve learned how to calculate the work done on an object by a constant force, but what about a force that changes as the object moves? Equation 9.14, the definition of work, is all we need:

\begin{matrix} W = \int_{s_{i}}^{s_{f}} F_{s} d s = area under the force-versus-position graph \\ (work done by a variable force) \end{matrix}

(9.21)

The integral sums up the small amounts of work $F_{s} d s$ done in each step along the trajectory. The only new feature, because $F_{s}$ now varies with position, is that we cannot take $F_{s}$ outside the integral. We must evaluate the integral either geometrically by finding the area under the curve (which we’ll do in the next example) or by actually doing the integration (which we’ll do in the next section).

EXAMPLE 9.7 Using work to find the speed of a car

A 1500 kg car is towed, starting from rest. FIGURE 9.16 shows the tension force in the tow rope as the car travels from $x = 0 m$ to $x = 200 m$ . What is the car’s speed after being pulled 200 m?

The figure illustrates a force-versus-position graph for a car.

FIGURE 9.16 Full Alternative Text

MODEL Let the system consist of only the car, which we model as a particle. We’ll neglect rolling friction. Two vertical forces, the normal force and gravity, are perpendicular to the motion and thus do no work.

SOLVE We can solve this problem with the energy principle, $Δ K = K_{f} - K_{i} = W$ . Here W is the work done by the tension force, but the force is not constant so we have to use the full definition of work as an integral. In this case, we can do the integral graphically:

\begin{array}{l} W & = & \int_{0 m}^{200 m} T_{x} d x \\ = & area under the force curve from 0 m to 200 m \\ = & \frac{1}{2} (5000 N) (200 m) = 500, 000 J \end{array}

The initial kinetic energy is zero, so the final kinetic energy is simply the energy transferred to the system by the work of the tension: $K_{f} = W = 500, 000 J$ . Then, from the definition of kinetic energy,

v_{f} = \sqrt{\frac{2 K_{f}}{m}} = \sqrt{\frac{2 (500, 000 J)}{1500 kg}} = 26 m / s

REVIEW $26 m/s \approx 55 mph$ is a reasonable final speed after being towed 200 m.