Suppose that an object is moving along the x-axis with an initial velocity \(v_i\) (which could be \(0m/s\) but, in general, can be any arbitrary value) and is displaced by an amount \(Δx\). Let's suppose that this object is acted upon by any arbitrary force \(\vec{F}\) as it is being displaced as illustrated in Figure 1. The work \(W\) done by \(\vec{F}\) is, by deffinition, given by

$$W≡\int_c\vec{F}·d\vec{R}.\tag{1}$$

For an object whose motion is restricted to the x-axis, Equation (1) simplifies to

$$W=\int{\vec{F}·d\vec{x}}.\tag{2}$$

Since the vertical component of the force \(\vec{F}_y\) is always perpendicular to \(d\vec{x}\) (the object's displacement), it follows that \(\vec{F}_y\) does not contribute to the work done on the object. Only \(\vec{F}_x\) contributes to the work done and thus Equation (2) simplifies to

$$W=\int{\vec{F}_x·d\vec{x}}.\tag{3}$$

Since \(\vec{F}_x\) is always parallel (or anti-parallel) to \(d\vec{x}\), we can simplify Equation (3) to

$$W=\int{F_xdx},$$

if \(\vec{F}_x\) is pointing in the same direction as \(d\vec{x}\). (If \(\vec{F}_x\) is pointing in the opposite direction to \(dx\), we can just add a minus sign in front of the integral.) We'll restrict our attention to problems in which the force applied to the object is constant. Given this assumption, we can further simplify the integral to

$$W=F_x\int{dx}=F_xΔx.\tag{4}$$

Since \(F_x\) is constant, we can simplify Newton's second law (so that it is expressed in terms of accelleration instead of the time rate-of-change of momentum) and substitute \(ma_x\) in place of \(F_x\) to get

$$W=ma_xΔx.\tag{5}$$

Also, recall that the equations of kinematics apply to any object moving at a constant acelleration. Since we have chosen to restrict our attention to objects acted upon by a constant force, the x-component of accelleration \(a_x\) must also be constant. This means that we can use the following equation from kinematics (in fact, we could use any of the equations from kinematics since they all apply to any object moving with a constant acelleration) to describe the object's motion:

$$v_f^2=v_i^2+2a_xΔx.$$

If we algebraically manipulate the above equation in order to solve for \(Δx\), we get

$$Δx=\frac{v_f^2-v_i^2}{2a_x}.$$

Substituting this result into Equation (5), we have

$$W=ma_x(\frac{v_f^2-v_i^2}{2a_x})=m(\frac{v_f^2-v_i^2}{2})=\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2=ΔKE.\tag{6}$$

What we have essentially just proved to ourselves is that there is some relationship between the work done on an object and that object's change in energy. Or to be more technical, we've shown that work is an energy transfer mechanism: a way of transfering energy into and out of an object. In this proof in particular, we've shown that whenever a force \(\vec{F}\) acts on an object as that object is being displaced but whose altitude (or height) isn't changing, then that force \(\vec{F}\) transfers kinetic energy into or out of the object (depending on whether \(W=ΔKE\) is positive or negative). Equation (6) is called the *work kinetic-energy theorem*. In the next lesson, we'll prove that if an object (initially at a height \(y_i\)) moves along any trajectory until it reaches a different height \(y_f\), the force of gravity \(\vec{F}_g\) transfers energy into or out of the object. But it *does not* transfer kinetic energy into or out of the object; instead, it only transfers potential energy into or out of the object.