Figure 7-19 gives the xcomponent ${\mathit{f}}_{\mathbf{x}}$of a force that can act on a particle. If the particle begins at rest at $\mathit{x}\mathit{=}\mathit{0}$, what is its coordinate when it has (a) its greatest kinetic energy, (b) its greatest speed, and (c) zero speed? (d) What is the particle’s direction of travel after it reaches $\mathit{x}\mathit{=}\mathbf{6}\mathbf{}\mathbf{m}$?

1. The graph of the x-component of force versus displacement along the x-axis acting on the particle is given.
2. The particle begins at rest from x=0 m.

We use work in terms of force and displacement and the work-energy principle. From the area under the curve, we get the work done, which we can consider as the change in kinetic energy of the particle. Using that, we can determine the coordinates of the particle at given points.

Formulae:

The work done on a particle due to a change in kinetic energy,

$\mathbf{W}\mathbf{=}\mathbf{∆}\mathbf{KE}$ (1)

The kinetic energy of a body,

$\mathbf{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{mv}}^{\mathbf{2}}$ (2)

We can use the work-energy principle.

The area under the curve in the graph gives the work done, so when work is positive, there is the greatest kinetic energy that can be said using equation (1).

Hence, from x=0 m to x=3 m work is positive, so the greatest kinetic energy is at 3 m.

From the above part, we got the greatest kinetic energy at x=3 m, so the greatest speed using equation (2) of the particle is at 3 m.

From the graph, the speed of the particle is zero at x=0, and the area of the graph from x=0 m to x = 6 m will give work done as zero. That means the change in kinetic energy using equation (1) is zero.

Therefore, at 6 m the speed of the particle is zero.

At x=6 m , the particle is at rest, but still, the force is acting in a negative direction.

Therefore, the direction of the particle will be in the direction of force, that is, in the negative direction.