## Illustration 6.3: Force and Displacement

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Two forces that we often use as examples when talking about work are the force of gravity and the elastic force of springs. We know from earlier chapters that the character of the two forces is different. The gravitational force on an object is always mg, while the spring force is dependent on how much the spring is stretched or compressed from equilibrium. As a consequence, the form for the work done by each force will be different. Restart.

In general, for a constant force, WORK = **F** · Δ**x** = F Δx cos(θ), where **F** is the constant force and Δ**x** is the displacement. F and Δx are the magnitude of the vectors, respectively.

The graph shows Fcos(θ) vs. distance for a 1-kg object near the surface of Earth **(position is given in meters)**. By checking the box you make the graph represent the Fcos(θ) vs. distance graph for a mass on a spring with k = 2 N/m (the equilibrium point of the spring is conveniently set at x = 0 m). You can enter in values for the starting and stopping points for the calculation of the work and then click the "evaluate area (integral)" button to calculate the work.

Begin by looking at the Fcos(θ) vs. distance graph for gravity (F_{y} vs. y). Gravity is a constant force (near the surface of Earth). Therefore, the magnitude of work done by gravity will be | mg Δy |. So, consider a ball at y = 0 m that drops to y = -2 m. Is the work done by gravity positive or negative? Use the graph to calculate the work. It is indeed positive (a negative force in the y direction and a negative displacement in the y direction means cos(θ) = 1). This is because the force is in the same direction as the displacement. What about lifting an object up from y = -2 m to y = 0 m? The work done is negative since the force is in the opposite direction from the displacement [cos(θ) = -1]. We can use | F Δy | because the force does not vary over the displacement. But what if the force does vary, as in the case of a spring?

Check the check box to see the graph representing a spring force. Enter in values for the starting and stopping points for the calculation of the work and then click the "evaluate area (integral)" button. Enter in x = 0 m for the starting point and x = 4 m for the ending point, representing the stretching of a spring. Is the magnitude of the work done | F Δx |? Why or why not? The magnitude of the work is not | F Δx |. In the case of the spring, the magnitude of the work is 0.5*k x^{2}, which is the area under the force function (it is also the integral of F dx). Note also that the work is negative: The force and the displacement are in the opposite direction [cos(θ) = -1].

Enter in x = 4 m for the starting point and x = 0 m for the ending point. What happens to the sign of the work done by the spring now?