B

Step 1: According to the principles of fluid mechanics, the relationship between shear stress ($\tau$) and shear rate ($\dot{\gamma}$) for a Newtonian fluid is given by: $\tau = \mu \dot{\gamma}$, where $\mu$ is the dynamic viscosity.

Step 2: The shear stress ($\tau$) can be defined as: $\tau = \frac{F}{A}$, where $F$ is the force applied and $A$ is the area over which it is applied.

Step 3: The shear rate ($\dot{\gamma}$) is defined as the velocity gradient. In our case, since the sleeve attains a speed of 2 m/s with a force of 788 N, the following relation holds true:
$$\dot{\gamma_1} = \frac{v}{d} = \frac{2}{d}$$ for force $F_1 = 788$ N.

Step 4: Let’s denote the new applied force as $F_2 = 1400$ N. We can write the relationship of shear stress for the two cases as:
$\frac{F_2}{A} = \mu \dot{\gamma_2}$
Given that the temperature remains constant, viscosity $\mu$ does not change.

Step 5: The velocities attained will therefore be related by:
$$\frac{788}{F_2} = \frac{\dot{\gamma_1}}{\dot{\gamma_2}}$$ where $\dot{\gamma_2} = \frac{v_2}{d}$$.

Step 6: Solving for $v_2$ when $F_2 = 1400$ N:
$$\frac{788}{1400} = \frac{2/d}{v_2/d}$$ This simplifies to:
$$v_2 = 2 \times \frac{1400}{788}$$.

Step 7: Calculating $v_2$:
$$v_2 = 2 \times 1.778 = 3.556\, m/s$$.
Rounding this, the speed of the sleeve would be approximately 3.56 m/s. Thus, the correct answer choice considering the options should be B.