A block of mass m is lying at  $x=0$ on a smooth horizontal surface. A variable force $F=kx$ applied to it as shown in figure where $k$ is constant. Then

The horizontal and vertical components of F are F cos $\text{θ}$ and F sin $\text{θ}$(see fig. 5.102)

$\text{N+F sin }\theta \text{\hspace{0.17em}=\hspace{0.17em}mg}$

$\Rightarrow N=mg-F\sin \theta$      

$=mg-kx\sin \theta$   

 Where N is the normal reaction The block will lose contact with surface at ${\text{x=x}}_0$ for which N=0 

putting N = 0 and x=x₀, we have

0 = mg - kx₀sin$\theta$ 

$\Rightarrow$ $x_0=\frac{mg}{k\sin \theta }$

The acceleration a of block along the horizontal surface is given by

Ma = Fcos$\theta$ = kxcos$\theta$

$\Rightarrow a=\frac{kx\cos \theta }{m}$

Which depends upon x. Hence the correct choices are (b) and (c)