A rough inclined plane is placed on a cart moving with a constant velocity u on horizontal ground. A block of mass M rests on the incline. Is any work done by force of friction between the block and incline ? Is there then a dissipation of energy ?
Shouldn”t force of friction between incline plane and block be zero? Since there is no relative motion between them .
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Can static friction dissipate energy ?
I like this question because it really makes you think.
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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:$$N sin heta = f cos heta \N cos heta + f sin heta = mg$$where $ heta$ 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 $f = mg sin heta$.)
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 $x$:$${
m work} = f cos ( heta) , x$$but that energy is immediately sent back to the inclined plane where it came from, via the normal reaction force:$${
m work} = N sin ( heta) , x .$$These two match, so neither entity either gains or loses energy overall.
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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.
