Work equals force times the component of displacement parallel to the force Steps: - Identify the formula for work by a constant force: $W = \vec{F} \cdot \Delta \vec{r} = F \Delta r \cos \theta$, where $\theta$ is the angle between force and displacement vectors. - Note the force magnitude $F = 2.1$ N and displacement magnitude $\Delta r = 4.0$ m. - Since no diagram is provided to determine $\theta$, the angle between the horizontal force and the displacement direction is unknown. - Without $\theta$ or additional details (e.g., initial velocity direction or time), the parallel component cannot be calculated numerically. Not enough information. Why C is correct: - C (7.3 J) would require $\cos \theta \approx 0.87$ ($\theta \approx 30^\circ$), likely from a diagram showing the force at an angle to the initial motion direction. Why the others are wrong: - A: Assumes displacement component of 2 m ($2.1 \times 2 = 4.2$), too small. - B: Assumes component of ~2.3 m, arbitrary and unsupported. - D: Assumes $\theta = 0^\circ$ (force …