Why is the work zero when you carry something at constant height

Work is defined as the scalar product of a constant force and the displacement of its point of application: (W = mathbf{F}cdotmathbf{d}=Fdcos\theta). The angle $\theta$ is measured between the direction of the force and the direction of motion. If the force is perpendicular to the motion, $\cos 90^{\circ}=0$ and the work done is exactly zero.

Carrying an object at constant height

When you lift a box onto a table and then walk across the room, you apply an upward force equal to the weight of the box, (F = mg). Your hand moves the box horizontally, so the displacement vector lies in the horizontal plane. Because the upward force has no horizontal component, the angle between $\mathbf{F}$ and $\mathbf{d}$ is 90°, and the work contributed by that force is zero.

Calculate work for a simple carrying scenario:

1. 1Identify the constant force: (F = mg = 10\,$\text{kg}$\times9.8\,$\text{m/s}$^2 = 98\,$\text{N}$\) upward.
2. 2Measure the horizontal displacement: (d = 5\,$\text{m}$\) east.
3. 3Find the angle: $\theta = 90^{\circ}$.
4. 4Compute work: (W = 98\times5\times\cos90^{$\circ$}=0\,$\text{J}$\).

Four situations that guarantee zero work:

• Force is perpendicular to displacement.
• Force is present but the point of application does not move.
• Displacement is zero (object remains stationary).
• Force is balanced by an equal opposite force, resulting in no net motion.

Comparison of force components during carrying: