A uniform solid cylinder of mass 2 kg rests on a long plank of mass 3 kg which lies on a smooth horizontal floor. A constant horizontal force of 22 N…

Let the plank accelerate at $a_p$ and the cylinder centre at $a_c$, both in the direction of the force. Friction $f$ on the cylinder acts forward, so $M a_c=f$ and its torque gives $fR=\tfrac12MR^2\alpha$, i.e. $\alpha=2a_c/R$. The contact point of the cylinder has no acceleration relative to the plank, so $a_p=a_c+R\alpha=3a_c$. For the plank, $F-f=m a_p$, giving $F=Ma_c+3ma_c$ and $a_c=F/(M+3m)=22/11=2\ \mathrm{m\,s^{-2}}$. Hence $a_p=6\ \mathrm{m\,s^{-2}}$ and $f=4\ \mathrm{N}$. The value $4.4$ results from treating plank and cylinder as one rigid body, $2.0$ is the cylinder's acceleration, and $7.3$ is $F/m$ for a frictionless contact.