Work done by friction on an inclined plane

Your thought about the inclined plane moving on the cart would be correct if there was no friction between the cart and the inclined plane.

However, if the whole system moves at velocity of $u$ there must have been some force that held everything together to allow for the cart to accelerate up to this velocity (otherwise the inclined plane would slide back). To visualize this, think about suddenly moving a cart when you go shopping. The items inside the cart slide back if the cart is pushed with too much force. This occurs because the static friction force cannot keep up with the amount of force that is required to allow the objects to accelerate with the cart. Thus, we know that there must be some friction between the cart and inclined plane.

Now, the friction between the inclined plane and cart cancels out the frictional force between the block and inclined plane that tries to move the inclined plane back.

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.