\begin{problem*}{7.28}
An $m_1 = 50.0\U{kg}$ block and an $m_2 = 100\U{kg}$ block connected by a string as shown in Figure P7.28.
The pulley is frictionless and of negligible mass.
The coefficient of kinetic friction between $m_1$ and the incline is $\mu = 0.250$.
The incline is at an angle of $\theta = 37.0\dg$ from the horizontal.
Determine the change in the kinetic energy of $m_1$ as it moces from point $A$ to point $B$, a distance of $d = 20.0\U{m}$.
\end{problem*} % problem 7.28

\begin{solution}
Again, we use conservation of energy.  Defining our gravitational
potential energy to be zero at $A$ we have
\begin{equation}
 E_A + W_f = K_{1A} + K_{2A} + W_f = E_B = K_{1B} + K_{2B} + U_{g1B} + U_{g2B}
\end{equation}
The blocks are tied together, so they must have the same velocity
(since the string remains taught).  So the change in velocity $v$ is
given by
\begin{align}
 \frac{1}{2}(m_1 + m_2) v_A^2 - F_f d &= \frac{1}{2}(m_1 + m_2) v_B^2 + m_1 g d \sin \theta - m_2 g d \\
 \Delta(v^2) = v_B^2 - v_A^2 &= \frac{2}{m_1+m_2} \cdot \left[ gd \cdot (m_2 - m_1 \sin\theta) - F_f d \right] \\
\end{align}
So the change in kinetic enery of $m_1$ is given by
\begin{equation}
 \Delta(K_1) = \frac{d m_1}{m_1+m_2} \cdot \left[ g (m_2 - m_1 \sin\theta) - F_f \right]
\end{equation}

We still need to find the force of fiction, which we do by
constructing a free body diagram of $m_1$.  We see that the forces on
$m_1$ are friction $\vect{F}_f$, tension \vect{T}, normal
$\vect{F}_N$, and gravitational $\vect{F}_{g1}$.  Summing the forces
in the direction perpendicular to the incline (\vect{y}), we have
\begin{align}
 \sum F_y &= F_N - F_{g1} \cos \theta = 0 \\
 F_N &= m_1 g \cos\theta
\end{align}
The block is always sliding so $F_f = \mu F_N = \mu m_1 g \cos\theta$.
Plugging this into our equation for $\Delta(K_1)$ we have
\begin{align}
 \Delta(K_1) &= \frac{d g m_1}{m_1+m_2} \cdot \left( m_2 - m_1 \sin\theta - \mu m_1 \cos \theta \right) \\
             &= \frac{20.0\U{m} \cdot 9.8\U{m/s}^2 \cdot 50.0\U{kg}}{150\U{kg}} \cdot \left[ 100\U{kg} - 50\U{kg}( \sin 27.0\dg - 0.250 \cos 27.0\dg ) \right] \\
             &= \ans{5.78\U{kJ}}
\end{align}
\end{solution}
