1. Given data

• Force applied tangentially at the highest point: $F = 49\,\text{N}$
• Mass of solid sphere: $m = 20\,\text{kg}$
• For a solid sphere, $I_{\text{cm}} = \frac{2}{5}mR^2$

We need the acceleration of the center when the sphere rolls without slipping.

2. Choose directions

Let rightward linear acceleration be $a$. Let clockwise angular acceleration be $\alpha$.

The force $F$ acts at the top tangentially toward the right. This force tends to:

• translate the sphere to the right,
• rotate it clockwise.

Let friction at the ground be $f$. We will determine its direction by equations.

3. Translational equation

Horizontal forces on the sphere:

• applied force $F$ to the right,
• friction $f$ at contact.

So, $F + f = ma \quad \text{(1)}$

4. Rotational equation about the center

Take clockwise torque as positive.

• Torque due to force at top: $FR$ clockwise
• Torque due to friction at bottom: friction to the right would produce anticlockwise torque, so torque is $-fR$

Hence, $FR - fR = I\alpha$ $R(F-f)=\frac{2}{5}mR^2\alpha$ $F-f=\frac{2}{5}mR\alpha \quad \text{(2)}$

5. Rolling without slipping condition

For pure rolling, $a = \alpha R$ So, $\alpha = \frac{a}{R}$

Substitute into (2): $F-f=\frac{2}{5}mR\left(\frac{a}{R}\right)=\frac{2}{5}ma \quad \text{(3)}$

6. Solve equations

From (1): $F+f=ma$

From (3): $F-f=\frac{2}{5}ma$

Add the two equations: $2F = ma + \frac{2}{5}ma = \frac{7}{5}ma$

Thus, $a = \frac{2F\cdot 5}{7m} = \frac{10F}{7m}$

Substitute $F=49\,\text{N}$ and $m=20\,\text{kg}$: $a = \frac{10\times 49}{7\times 20}$ $a = \frac{490}{140} = 3.5\,\text{m/s}^2$

7. Check option

$a = 3.5\,\text{m/s}^2$

So the correct option is B.

8. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.