A force of 49 N acts tangentially at the highest point of a sphere (solid) of mass 20 kg, kept on a rough horizontal plane. If the sphere rolls without slipping, then the acceleration of the center of the sphere is

The force F acts at a distance of 2R from point P, and for a solid sphere, the moment of inertia about the contact point (using Parallel Axis Theorem) is:

$$I_P = I_{cm} + MR^2 = \frac{2}{5}MR^2 + MR^2 = \frac{7}{5}MR^2$$

For rolling without slipping, we can take torque about the point of contact (P) on the ground:

$$\tau_P = I_P \alpha$$

$$F(2R) = \left( \frac{7}{5}MR^2 \right) \frac{a}{R}$$

$$a = \frac{10F}{7M}$$

$$a = \frac{10 \times 49}{7 \times 20}$$

$$a = \frac{490}{140} = 3.5 \text{ m/s}^2$$