A uniform solid cylindrical roller of mass m is being pulled on horizontal surface with force F parallel to…

By Newton's law of motion

$F-f_s=ma$                   ...(i)

and torque, $\tau =l\alpha$

where, $\alpha =$ tangential component of acceleration
$I=\text{ moment of inertia }$

$\Rightarrow fR=l\alpha$

$fR=\frac{M{R}^2}{2}·\frac{a}{R}\left[\therefore I=\frac{M{R}^2}{2}\right.$and $\left.,\alpha =\frac{a}{R}\right]$

$\Rightarrow f=\frac{ma}{2}$

$\therefore F=f+ma$

$=\frac{ma}{2}+ma=\frac{3ma}{2}$