3.3 Work-Energy Theorem

As a force does work on an object, its speed will increase or decrease. This relationship is clarified by the work-energy theorem.

Theorem 3.1 (Work-Energy Theorem)

The net work on an object is equal to the change in its kinetic energy.

$\displaystyle W$	$\displaystyle=\int_{\mathrm{path}}\mathrm{d}W$	(3.36)
	$\displaystyle=\int_{r_{1}}^{r_{2}}\boldsymbol{F}\cdot\mathrm{d}\boldsymbol{r}$	(3.37)
	$\displaystyle=\int_{t_{1}}^{t_{2}}\boldsymbol{F}\cdot\frac{\mathrm{d}% \boldsymbol{r}}{\mathrm{d}t}\mathrm{d}t$	(3.38)
	$\displaystyle=\int_{t_{1}}^{t_{2}}\boldsymbol{F}\cdot\boldsymbol{v}\mathrm{d}t$	(3.39)
	$\displaystyle=\int_{t_{1}}^{t_{2}}\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{1}% {2}mv^{2}\right)\mathrm{d}t=\Delta K.$	(3.40)

3.3.1 Using the Work-Energy Theorem

The kinetic energy depends on the speed of the object, so if the net work is $>0$ then the object must have sped up. Likewise, if the net work is $<0$ , the object has slowed down. If the net work is 0, the object must be at the same speed that it started at.

Example 3.6.

Consider a particle of mass $2\text{\,}\mathrm{kg}$ . It is being acted on by a force with the form $\boldsymbol{F}=4x\hat{\boldsymbol{\imath}}$ , and at $t=0$ its position is $x=$4\text{\,}\mathrm{m}$$ and its velocity is $v_{0}=$1\text{\,}\mathrm{m}\text{\,}{\mathrm{s}}^{-1}$$ . What is the particle’s velocity when it reaches $x=$6\text{\,}\mathrm{m}$$ ?

We can solve this using the work energy theorem. First we calculate the work done:

W=\int_{x_{1}}^{x_{2}}F_{x}\operatorname{\mathrm{d}}\!x=\int_{4}^{6}4x% \operatorname{\mathrm{d}}\!x=\left.2x^{2}\right|_{4}^{6}=$40\text{\,}\mathrm{J% }$.

(3.41)

Now we can use the fact that the work done is equal to the change in kinetic energy to get a formula for the final velocity:

$\displaystyle W$	$\displaystyle=\Delta K=\frac{1}{2}m(v_{f}^{2}-v_{0}^{2})$	(3.42)
$\displaystyle\implies v_{f}$	$\displaystyle=\sqrt{\frac{2W}{m}+v_{0}^{2}}$	(3.43)
	$\displaystyle=$6.4\text{\,}\mathrm{m}\text{\,}{\mathrm{s}}^{-1}$.$	(3.44)

3.3.2 Potential Energy for a Spring

Now let’s look at a mass on a spring. The restoring force on the mass always acts opposite to displacement. Hooke’s law says

F_{s}=-k(x-x_{0}),

(3.45)

where $k$ is the spring constant, $x_{0}$ is the equilibrium position of the spring. Then the work done by the spring force is

	$\displaystyle W_{s}=\int_{x_{0}}^{x_{f}}F_{s}(x)\operatorname{\mathrm{d}}\!x$	$\displaystyle=-\int_{x_{0}}^{x_{f}}k(x-x_{0})\operatorname{\mathrm{d}}\!x$		(3.46)
		$\displaystyle=-\frac{1}{2}k(x_{f}-x_{0})^{2}.$		(3.47)

Note that the work done is always negative no matter if the displacement is positive or negative because the force points in the opposite direction. Now we define the spring potential energy as

U_{s}=-W_{s}=\frac{1}{2}k(x_{f}-x_{0})^{2},

(3.48)

Then by the work-energy theorem we have

\Delta K=W_{s}=-\Delta U_{s},

(3.49)

K+U_{s}=E

(3.50)

is constant.