In the following diagram, is work done by static friction 0 ?, since the point of application is also moving with speed v w.r.t. ground here and is only stationary w.r.t. the block on which sphere is rolling w.r.t. ground here.
Static friction itself is 0. The formula $f_s=\mu N$ defines the maximum possible magnitude of the static friction force, not the true static friction force. In this case, there is no other acceleration, so there is no need for static friction. Static friction only comes into play when the two bodies are attempting to be in relative motion with each other. This is not the case here, at the point of contact the velocities of the corresponding points on the wheel and platform are equal and there is no force trying to stop this.
When you're standing on the ground, you're not mysteriously being pushed by friction. It's the same thing here, the wheel is "standing" with respect to the point of contact, though the points of contact are changing over time.
Just addressing the question in this comment about banked curves. Static friction is always going to oppose the motion that would happen if there were no friction. I will use the free-body diagram here as a reference for the case of no friction. The only two forces on the car are the normal force (N) and gravity (mg). The sum of these two forces is in the horizontal direction toward the center of the circle that the car is traveling around. This net force is what keeps the car traveling in a circle, and is equal to a component of the normal force. Now, if we consider the fourth equation on that page, which comes from considering $F_{net}=F_{centripetal}$:
$$mg\tan{\theta} = \frac{mv^2}{r}$$
And divide by $m$:
$$g\tan{\theta} = \frac{v^2}{r}$$
This equation says for the car to stay in uniform circular motion (speed $v$ and radius $r$ don't change), there must be a balance between the four parameters in this equation. If, for example, speed $v$ is increased, radius $r$ must also increase given that $g$ and $\theta$ are constant. In the case that the car starts increasing its speed, it will start to slide up the incline. In this case, it will do so, and stop sliding sideways once the equation above is satisfied.
However, if we consider the case of an incline with friction, the situation changes. First, if the equation above is satisfied, then no friction will act sideways on the car tires (it isn't necessary, the car isn't trying to move sideways). However, there will be some friction on the tires in the forward direction, as in any case of "rolling without slipping." That phrase means the point of the tire that is in contact with the road at any instant in time is not moving w.r.t the road. It is "trying" to move backwards (think about a car at rest on ice. If you try to accelerate too quickly, the tires spin, and the point on the ice moves backwards. It is the same with a car in motion.), so the force from static friction must push forwards on the tire. This is what allows the car to accelerate forwards.
In the same scenario (rolling without slipping, $g\tan{\theta} = \frac{v^2}{r}$ initially satisfied) if the car's speed increases, then the equation will no longer be satisfied. But there is friction now, and as we found in the frictionless case, the car will "try" to move up the incline, and thus static friction will point down to oppose this motion. Vice versa, if the car decreases its speed, static friction will point up to oppose the car "trying" to slide down. (In this way, you can drive at a range of speeds for a given $g$, $\theta$, and $r$ if static friction is present.)
So, static friction always opposes attempted motion between two surfaces in contact.
Best Answer
I like this question because it really makes you think.
First, draw a diagram showing all the forces on the block. There is force $mg$ owing to gravity, straight down; normal reaction force $N$ orthogonal to the plane; and static friction force $f$ along the plane. The block is not accelerating so all these are balanced: $$ N \sin \theta = f \cos \theta \\ N \cos \theta + f \sin \theta = mg $$ where $\theta$ is the angle of the incline. So for your answer, the main point so far is that the friction force is not zero. (You get $f = mg \sin \theta$.)
Now is this force doing any work? That it is the puzzle. The thing it is acting on is in motion, with a component of velocity in the direction of the force, therefore the friction force is indeed doing work. But no energies are changing here, so how can that be? The answer is that the normal reaction force on the block is also doing work, and these two amounts of work exactly balance out. The total force on the block here is zero, so does no work. But each force which has a non-zero component in the direction of motion of the block does do some work. The friction between plane and block is providing energy to the block as it moves through each small distance $x$: $$ {\rm work} = f \cos (\theta) \, x $$ but that energy is immediately sent back to the inclined plane where it came from, via the normal reaction force: $$ {\rm work} = N \sin (\theta) \, x . $$ These two match, so neither entity either gains or loses energy overall.
Examples like this are using the concept of work in a rather unusual way, however. I would rather look at the total force on each object, and asking whether that total force is doing any work. Clearly if the total force is zero then it cannot be doing any work. Looking at the various parts of the total, as I have done here, is all correct but it is a little unusual. You rarely need to do this in order to understand some physical problem, until you look at the mechanics of continuous media such as fluids and vibrating solids, but that is a more advanced subject.