Substitute $N = Mg \cos\theta$ into equation (1) to find the friction force $f$ required to maintain the wall force $F$: $$ (Mg \cos\theta) \sin\theta – f \cos\theta = F $$ $$ f \cos\theta = Mg \sin\theta \cos\theta – F $$ $$ f = Mg \sin\theta – \frac{F}{\cos\theta} $$

Now substitute this expression for $f$ into the wheel’s equation of motion: $$ Mg \sin\theta – \left( Mg \sin\theta – \frac{F}{\cos\theta} \right) = Ma_{cm} $$ $$ \frac{F}{\cos\theta} = Ma_{cm} $$ $$ a_{cm} = \frac{F}{M \cos\theta} $$

The acceleration $a_{cm}$ is constant and depends only on the given parameters ($F, M, \theta$). Using kinematics equation $v^2 = u^2 + 2as$ with $u=0$: $$ v^2 = 2 \left( \frac{F}{M \cos\theta} \right) s $$ $$ v = \sqrt{\frac{2Fs}{M \cos\theta}} $$ This formula is valid for both rolling with slipping and rolling without slipping, as it is derived directly from the force constraint $F$.

We must compare the speed acquired when rolling with slipping ($v_{slip}$) versus rolling without slipping ($v_{roll}$).
While the formula derived above relates speed to the measured force $F$, physically, a wheel slides faster than it rolls because less energy is diverted into rotation (or alternatively, friction is lower during slipping than the static friction required to induce pure rolling).
Therefore, for the same physical setup (where $F$ would consequently be different), the wheel that slips accelerates faster.