\begin{problem*}{6.24}
An $m = 4.00\U{kg}$ particle is subject to a total force that varies
with position as shown in Fig. P6.11.  The particle starts from rest
at $x = 0$.  What is the speed at
 \Part{a} $x = 5.00\U{m}$,
 \Part{b} $x = 10.0\U{m}$, and
 \Part{c} $x = 15.0\U{m}$?
\end{problem*} % problem 6.24

\begin{solution}
In each of these cases we'll be conserving energy.  The energy put in
by the force all goes into the particle's kinetic energy.  So
conserving energy we have
\begin{align}
 W_{0,5} &= \int_{0\U{m}}^{5\U{m}} \vect{F} \cdot \vect{dx}
          = \frac{1}{2} m v^2 \\
 v_5 &= \sqrt{ \frac{2 W_{0,5} }{m} }
      = \sqrt{ \frac{2 \cdot 1.5\U{N} \cdot 5\U{m}}{4.00\U{kg}} }
      = \ans{1.94\U{m/s}}
\end{align}
Doing the same for the other works
\begin{align}
 W_{0,10} &= W_{0,5} + 3\U{N} \cdot 5\U{m} = 22.5\U{J} \\
 W_{0,15} &= W_{0,10} + 1.5\U{N} \cdot 5\U{m} = 30\U{J}
\end{align}
So the other velocities are
\begin{align}
 v_{10} &= \sqrt{ \frac{2 W_{0,10} }{m} } = \ans{3.35\U{m/s}} \\
 v_{15} &= \sqrt{ \frac{2 W_{0,15} }{m} } = \ans{3.87\U{m/s}}
\end{align}
\end{solution}
