Answer
The horizontal acceleration of the block has a magnitude of $1.83m/s^2$ and is directed leftward.
Work Step by Step
We take rightward to be $+x$ direction. First, we need to find the horizontal component of $F_1$ and $F_2$
$F_{1, x}=F_1\cos70=59\cos70=20.18N$
Since $F_2$ is already a horizontal force which points in $-x$ direction, $F_{2,x}=F_2=-33N$
Therefore, $\sum F_x=F_{1, x}+F_{2, x}=-12.82N$
According to Newton's 2nd Law, $$\sum F_x=m_{block}a_x$$ $$a_x=\frac{\sum F_x}{m_{block}}=\frac{-12.82}{7}=-1.83m/s^2$$
- Magnitude: $a_x=1.83m/s^2$
- Direction: $a_x\lt0$ means that it is directed leftward.
