Answer
$$\begin{aligned} a =1.3 \mathrm{\ m} / \mathrm{s}^{2} \end{aligned}$$
Work Step by Step
The second law equations for the moving crate are
$$
T \cos \theta-f=m a, \quad T \sin \theta+F_{N}-m g=0
$$
Now $f=\mu_{k} F_{N},$ and the second equation above gives $F_{N}=m g-T \sin \theta,$ which then yields $f=\mu_{k}(m g-T \sin \theta) .$ This expression is substituted for $f$ in the first equation to obtain
$$
\begin{array}{c}{T \cos \theta-\mu_{k}(m g-T \sin \theta)=m a}\end{array}
$$
so the acceleration is
$$\begin{aligned} a &=\frac{T\left(\cos \theta+\mu_{k} \sin \theta\right)}{m}-\mu_{k} g \\ &=\frac{(304 \mathrm{N})\left(\cos 15^{\circ}+0.35 \sin 15^{\circ}\right)}{68 \mathrm{kg}}-(0.35)\left(9.8 \mathrm{m} / \mathrm{s}^{2}\right)=1.3 \mathrm{\ m} / \mathrm{s}^{2} \end{aligned}$$
Let's check the limit where $\theta=0 .$ In this case, we recover the familiar expressions: $$T=\mu_{s} m g \quad \text{ and }\quad a=\left(T-\mu_{k} m g\right) / m $$
