Answer
a) $\rm N$
b) $\rm m^{-1}$
c) $\dfrac{\pi}{2c}$
d) See the figure below.
e) $\sqrt{ \dfrac{ 2F_0}{mc}+ v_0^2 }$
Work Step by Step
a) We know that $\sin\theta$ is dimensioless, so $\sin(cx)$ is also dimensioless. This means that the unit of $F_0$ is $\boxed{\bf Newton\;(N)}$.
___________________________________________
b) As we mentioned above, $(cx)$ is dimensionless. And since we know that the unit of $x$ is meter $\rm (m)$, thus the unit of $c$ must be $\boxed{\rm\bf m^{-1}}$
___________________________________________
c) $F_{max}$ occurs when $\sin(cx)=1$ which occurs when $cx=90^\circ$ or $=\pi/2$.
Thus,
$$cx_{max}=\dfrac{\pi}{2}$$
$$\boxed{x_{max}=\dfrac{\pi}{2c}}$$
___________________________________________
d) See the graph below.
___________________________________________
e) According to the Work-kinetic energy theorem,
$$W=\Delta K=K_f-K_i$$
Thus,
$$W=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2\tag 1$$
We need to find $v_f$ when $x=x_{max}$ which occurs when $F=F_0$.
The work done by the force is given by
$$W=\int_{x_0}^{x_{max}} Fd_x=\int_{x_0}^{x_{max}} F_0\sin(cx)d_x$$
$$W= F_0\int_{x_0}^{x_{max}} \sin(cx)d_x=F_0\left[\dfrac{-1}{c}\cos(cx)\right]\bigg|_{x_0}^{x_{max}}$$
$$W=\dfrac{-F_0}{c}\cos(cx)\bigg|_{x_0}^{x_{max}}$$
where $x_{max}=\pi/2c$ and $x_0=0$
$$W=\dfrac{-F_0}{c}\left[\cos\left(\color{red}{\bf\not} c\dfrac{\pi}{2\color{red}{\bf\not} c}\right)-\cos\left(c\times 0\right)\right]$$
$$W=\dfrac{-F_0}{c}\left[0-1\right]=\dfrac{ F_0}{c}$$
Plugging into (1);
$$\dfrac{ F_0}{c}=\frac{1}{2}mv_f^2-\frac{1}{2}mv_0^2 $$
$$\dfrac{ 2F_0}{c}= mv_f^2- mv_0^2 $$
$$\dfrac{ 2F_0}{c}+ mv_0^2 = mv_f^2$$
$$\dfrac{\dfrac{ 2F_0}{c}+ mv_0^2}{m} = v_f^2$$
$$\boxed{v_f=\sqrt{ \dfrac{ 2F_0}{mc}+ v_0^2 }}$$
