Answer
$a$ has a magnitude of $1.65m/s^2$ and is directed at an angle of $34.6^o$ relative to $+x$
Work Step by Step
Rewrite $F_1$ and $F_2$ in unit-vector notation:
- $\vec{F}_2=(54N)i$
- $\vec{F}_1=(88\cos55N)i+(88\sin55N)j=(50.47N)i+(72.09N)j$
Therefore, $\vec{F}=(104.47N)i+(72.09N)j$
In magnitude, $F=\sqrt{104.47^2+72.09^2}=126.93N$
$F$ is opposed by kinetic friction $f_k$, whose magnitude is $$f_k=\mu_kF_N=\mu_kmg=0.35\times25\times9.8=85.75N$$
Therefore, $\sum F=F-f_k=41.18N$
According to Newton's 2nd Law: $$a=\frac{\sum F}{m_{crate}}=\frac{41.18}{25}=1.65m/s^2$$
The direction of $a$ is similar with the direction of force $F$ that propels the motion.
So, if we call $\theta$ the angle of $\vec{a}$ relative to $+x$, we have $$\tan\theta=\frac{72.09}{104.47}=0.69$$ $$\theta=34.6^o$$
