1. Key idea: constant velocity implies zero tangential acceleration

Since the object moves with constant velocity, its acceleration is zero along the direction of motion. Hence, at every point on the frictionless track, the applied force must exactly balance the component of weight along the surface.

So, $F_{\text{applied}} = mg\sin\theta$ where $\theta$ is the instantaneous inclination of the surface with the horizontal.

2. Interpretation of the given statement

The problem says that the object is moved with constant velocity under an applied force of 2 N along the frictionless surface profile.

That means at those portions where the slope is constant and such that $mg\sin\theta = 2\,\text{N},$ the required applied force is $2\,\text{N}$.

3. How force changes with the profile

Because the surface is frictionless, the required applied force depends only on the local slope:

• Horizontal portion: $\theta = 0$ $F = mg\sin 0 = 0$
• Upward incline with constant slope: $F = mg\sin\theta = \text{constant positive}$
• Downward incline with constant slope: the component of gravity is along the motion, so to keep speed constant, the applied force must act opposite to motion. If we take forward direction as positive, then $F = -mg\sin\theta$
• Steeper incline $\Rightarrow$ larger magnitude of force.

4. Shape of the $F$ vs distance graph

Since the profile consists of piecewise straight segments, the inclination $\theta$ is piecewise constant. Therefore the applied force is also piecewise constant as a function of distance.

Thus the graph must:

• be zero on horizontal parts,
• be constant positive on upward slopes,
• be constant negative on downward slopes,
• switch values abruptly when the slope changes.

5. Matching with the options

Among the given choices, the only graph that correctly shows the applied force taking piecewise constant values corresponding to the changing slope of the surface profile is Option C.

Therefore, the correct answer is: $\boxed{C}$