A plank of mass 4.0 kg and length 1.5 m lies on a smooth horizontal floor…

Step 1. The equations. The plank drags the bottom of the cylinder forwards, so friction $f$ on the cylinder points forwards (and backwards on the plank).

• Cylinder, moving: $f = ma_c$.
• Cylinder, turning about its centre: $fR = I\alpha$ with $I = \tfrac12 mR^2$, so $\alpha R = \dfrac{2f}{m}$. This spin makes the bottom move forwards relative to the centre.
• Plank: $F - f = Ma_p$.

Step 2. Rolling without slipping. The bottom of the cylinder moves with the plank: $a_c + \alpha R = a_p$, so $\dfrac{f}{m} + \dfrac{2f}{m} = a_p$ and $f = \dfrac{ma_p}{3}$. Then $F - \dfrac{m}{3}a_p = Ma_p$:

So $a_c = \dfrac{a_p}{3} = 1.57\ \text{m/s}^2$ and $f = ma_c = 3.14\ \text{N}$. The cylinder moves forwards over the floor, but backwards relative to the plank.

Step 3. Least friction. Rolling needs $f \le \mu mg$: $\mu \ge \dfrac{3.143\ \text{N}}{20\ \text{N}} = 0.157$.

Step 4. With $\mu = 0.10$. This is less than $0.157$, so the cylinder slips. Sliding friction is $\mu mg = 2.0\ \text{N}$, still forwards on the cylinder:

Check the direction: the bottom of the cylinder speeds up at $a_c + \alpha R = 3.0\ \text{m/s}^2$, less than the plank's $5.0\ \text{m/s}^2$. So the plank keeps sliding forwards under it, and friction stays forwards.

Step 5. Time. Relative to the plank, the centre moves backwards with $5.0 - 1.0 = 4.0\ \text{m/s}^2$ from rest. It must go $1.5\ \text{m}$: $t = \sqrt{\dfrac{2 \times 1.5\ \text{m}}{4.0\ \text{m/s}^2}} = 0.87\ \text{s}$.