# 8.3: Work-Energy Principle

Difficulty Level: At Grade Created by: CK-12

The reason the concept of work is so useful is because of a theorem, called the work-energy principle, which states that the change in an object's kinetic energy is equal to the net work done on it:

ΔKe=Wnet [2]\begin{align*}\Delta K_e =W_{net} \text{ [2]}\end{align*}

Although we cannot derive this principle in general, we can do it for the case that interests us most: constant acceleration. In the following derivation, we assume that the force is along motion. This doesn't reduce the generality of the result, but makes the derivation more tractable because we don't need to worry about vectors or angles.

Recall that an object's kinetic energy is given by the formula:

Ke=12mv2 [3] \begin{align*}K_e = \tfrac{1}{2}mv^2 \text{ [3] }\end{align*}

Consider an object of mass m\begin{align*} m \end{align*} accelerated from a velocity vi\begin{align*} v_i \end{align*} to vf\begin{align*} v_f \end{align*} under a constant force. The change in kinetic energy, according to [2], is equal to:

ΔKe=KeiKef=12mv2f12mv2i=12m(v2fv2i) [4] \begin{align*}\Delta K_e = K_{ei}- K_{ef} = \tfrac{1}{2}mv_f^2-\tfrac{1}{2}mv_i^2 = \tfrac{1}{2}m(v_f^2-v_i^2) \text{ [4] }\end{align*}

Now let's see how much work this took. To find this, we need to find the distance such an object will travel under these conditions. We can do this by using the third of our 'Big three' equations, namely:

Undefined control sequence \intertext\begin{align*}{v_f}^2 = {v_i}^2 + 2a\Delta x \text{ [5]} \intertext{alternatively,} \Delta x = \frac{{v_f}^2-{v_i}^2}{2a} \text{ [6] }\end{align*}

Plugging in [6] and Newton's Third Law, F=ma\begin{align*}{F} = ma \end{align*}, into [2], we find:

W=FΔx=ma×vf2vi22a=12m(v2fv2i) [7],\begin{align*}W = F \Delta x = ma\times \frac{{v_f}^2-{v_i}^2}{2a} = \tfrac{1}{2}m(v_f^2-v_i^2)\text{ [7],}\end{align*}

which was our result in [4].

## Using the Work-Energy Principle

The Work-Energy Principle can be used to derive a variety of useful results. Consider, for instance, an object dropped a height Δh\begin{align*} \Delta h \end{align*} under the influence of gravity. This object will experience constant acceleration. Therefore, we can again use equation [6], substituting gravity for acceleration and Δh\begin{align*} \Delta h \end{align*} for distance:

Undefined control sequence \intertext\begin{align*}\Delta h &= \frac{{v_f}^2-{v_i}^2}{2g} \intertext{multiplying both sides by mg, we find:} mg \Delta h &= m \cancel{g} \frac{{v_f}^2-{v_i}^2}{2\cancel{g}} = \Delta K_e \text{ [8]}\end{align*}

In other words, the work performed on the object by gravity in this case is mgΔh\begin{align*}mg \Delta h\end{align*}. We refer to this quantity as gravitational potential energy; here, we have derived it as a function of height. For most forces (exceptions are friction, air resistance, and other forces that convert energy into heat), potential energy can be understood as the ability to perform work.

## Spring Force

A spring with spring constant k\begin{align*} k \end{align*} a distance Δx\begin{align*} \Delta x \end{align*} from equilibrium experiences a restorative force equal to:

Fs=kΔx [9]\begin{align*}F_s = -k\Delta x \text{ [9]}\end{align*}

This is a force that can change an object's kinetic energy, and therefore do work. So, it has a potential energy associated with it as well. This quantity is given by:

Esp=12kΔx2[10] Spring Potential Energy\begin{align*}E_{sp} = \tfrac{1}{2}k{\Delta x}^2 && \text{[10] Spring Potential Energy}\end{align*}

The derivation of [10] is left to the reader. Hint: find the average force an object experiences while moving from x=0\begin{align*} x = 0 \end{align*} to x=Δx\begin{align*} x = \Delta x \end{align*} while attached to a spring. The net work is then this force times the displacement. Since this quantity (work) must equal to the change in the object's kinetic energy, it is also equal to the potential energy of the spring. This derivation is very similar to the derivation of the kinematics equations --- look those up.

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