**How the work done depends on the angle \(θ\)**

The work \(W\) done on an object acted upon by a constant force \(\vec{F}\) is given by

$$W=\vec{F}·Δ\vec{R},\tag{1}$$

where \(Δ\vec{R}\) is the displacement of the object. As a reminder, the dot product between any two vectors \(\vec{A}\) and \(\vec{B}\) is defined as

$$\vec{A}·\vec{B}≡ABcosθ,$$

where \(θ\) is the angle between those two vectors. From this definition, you can see that if the two vectors \(\vec{A}\) and \(\vec{B}\) are perpendicular to each other, then \(cos(±90°)=0\) and, thus, \(\vec{A}·\vec{B}=0\).

This basically means that if an object moves from \(\vec{R}(t_0)\) to \(\vec{R}(t)\) with a displacement of \(Δ\vec{R}\), if a constant force \(\vec{F}\) was acting on that object during its displacement and if \(\vec{F}\) was perpendicular to \(Δ\vec{R}\), then that force would have done no work on the object. For example, the Moon revolves around the Earth in a roughly circular path and the force of gravity \(\vec{F}_g\) exerted on the Moon by the Earth is always towards the center of Earth. Since this force is always acting on the Moon in a direction perpendicular to its displacement, it follows that the force \(\vec{F}_g\) doesn't do any work on the Moon.

We can rewrite Equation (1) in the following way:

$$W=\vec{F}·Δ\vec{R}=FΔRcosθ=(Fcosθ)ΔR=F_RΔR,\tag{2}$$

where \(F_R\) is the component of the force \(\vec{F}\) which is acting in the same direction as \(Δ\vec{R}\). Equation (2) tells us that it is only the component of force that is parallel to \(Δ\vec{R}\) which does work on the object. Therefore, if I apply the force \(\vec{F}\) to a box moving to the right as in Figure 1, the y-component of force \(F_y\) will contribute zero work to the object. It is only the component of force \(F_x=Fcosθ\) which is doing any work on the box. By changing the angle \(θ\) at which \(\vec{F}\) (while keeping the magnitude \(F\) the same) is applied in a way that increases \(cosθ\) (in this example, this can be accomplished by rotating \(\vec{F}\) clockwise), the same force \(\vec{F}\) will do more work on the object.