A solid sphere of mass 0.50 kg rests on a horizontal surface. The...

Let the applied horizontal force at the highest point be $F$ and the force of static friction at the contact point be $f$, acting in the forward direction.

For translational motion, Newton's second law gives:
$F + f = ma$

For rotational motion about the center of mass, the torque equation is:
$F R - f R = I \alpha$

For a solid sphere, the moment of inertia is $I = \dfrac{2}{5} m R^2$. For pure rolling (no slipping), the acceleration of the center of mass is $a = \alpha R$.

Substituting these into the torque equation:
$F R - f R =$\left(\dfrac{2}{5}m R^2\right) \left(\dfrac{a}{R}\right)<br>F - f = $\dfrac{2}{5}$ma$<br><br>Now, we have a system of two equations:<br>1)$F + f = ma$<br>2)$F - f = $\dfrac{2}{5}$ma$<br><br>Adding the two equations:<br>$2F = $\dfrac{7}{5}$ma \Rightarrow ma = $\dfrac{10}{7}$F$<br><br>Substituting$ma$back into the first equation to find$f$:<br>$F + f = $\dfrac{10}{7}$F \Rightarrow f = $\dfrac{3}{7}$F$<br><br>To prevent slipping, the static friction must not exceed its maximum possible value:<br>$f \le \mu_s N$<br>\dfrac{3}{7}$F \le \mu_s mg$<br><br>Solving for$F_\max$:<br>$F_\max = $\dfrac{7}{3} \mu_s mg$

Given $m = 0.50 \text{ kg}$, $\mu_s =$\dfrac{2}{7}, and taking $g = 9.8 \text{ m/s}^2$:
F_\max =\dfrac{7}{3} \times \dfrac{2}{7} \times$0.50 \times 9.8$
F_\max =\dfrac{1}{3} \times$9.8 = 3.266... \text{ N}$

Rounding to two significant figures, we get F_\max \approx 3.3 \text{ N}.