When is the net work done by a force equal to zero? My question is regarding work by a force and it is one which is giving me many issues in understanding when work done by force is 0.
My question can be described as a multiple choice question. 

Q) No work is done by a force on an object if -
a) the force is always perpendicular to velocity
b) the force is always perpendicular to acceleration
c) the object is stationary but point of application of force moves on the object
d) the object moves in such a way that the point of application of force remains fixed

The answers are (a), (c) and (d)
Now obviously I know (a) is true. Can somebody please explain :


*

*What the OPTIONS (c) and (d) mean. I'm unable to understand the options clearly. What does THIS POINT OF APPLICATION OF FORCE REFER TO? Is it the force or the point on which force is acting?

*Relate (c) and (d) to a sphere pure rolling. I know in pure rolling velocity of point of contact is 0 and hence point of contact doesn't move and hence work done by friction is 0. Now how to analyse using (c) or (d)? Which option does sphere rolling satisfy? Is it (c) or (d)? 

*Relate (c) and (d) with a pulley of mass M and moment of Inertia I about axis of rotation assuming string doesn't slip on pulley.  I'm unable to analyse this situation in any way. How is work done by both tensions on a ROTATING PULLEY with mass 0? (The string doesn't slip on pulley and pulley has mass M and moment of inertia I about axis).Which option does pulley rotating satisfy? Is it (c) or (d)? 

*Examples for (c) and (d). I want to first understand (c) and (d) and then relate it with many other examples
I'd really like a SIMPLE explanation highlighting the above issues in order to analyse WHEN WORK DONE BY A FORCE IS 0 without any problems.
 A: Option c is correct as, even though there are many forces the displacement due to every force is 0,therefore, summation(f.ds)=0 . 
U cannot relate it to rolling case as there are many forces acting but on rolling there is friction only.
For opt d, u can see it as the body divided into 2 small pieces,namely
1 The point of application piece
2  the rest of the body piece. 
So since work is  a scalar quantity we can add the work done on the two pieces after calculating them separately. 
Note: force acts only on  piece 1 and not on piece 2.
Work no
1.On point of application piece:F*(displacement) =F*0=0


*On other pieces : F*(d) =0*(d)=0. 


Work total=W1+W2=0+0=0.
Hence ur answer. 
U can relate it to the rolling problem, the point of applications of force of friction is the same i. e. the lower most point,
Hence the work done by friction is zero. 
A: The net work done by a force is equal to zero... when the net work is zero, of course! Let's first define what the net work is, then try to answer this question. Work is defined as the inner product of force and distance:
$\mathbf{W} = \mathbf{F} \bullet \mathbf{d}$
Note that these are vector quantities, so the line of force and line of direction must match for any work to occur. This means that we have a zero net force when one of two things happen:
1) There is net zero force along the direction $\mathbf{d}$.
or (that is an inclusive or, meaning one or both of these may be true)
2) There is net zero distance along which the force ($\mathbf{F}$) is applied.
How do we know what the force is? Newton's law of motion:
$\mathbf{F} = m\mathbf{a}$.
I'll give two examples of when zero net work may occur:
1) A circular orbiting satellite. A circular orbiting satellite, moving in a circle, is always accelerating towards the center of the orbit (let's say Earth). The direction of acceleration, and force (see Newton's equation), happen to be perpendicular to the velocity (i.e. motion, or distance traveled). Therefore, the gravitational force (which is the force acting on the satellite) is doing zero net work on the satellite because the satellite is not moving along the direction of the force (they are perpendicular).
2) A stationary building in strong winds. Strong winds acting on a stationary building perform zero net work on the building, no matter where the wind blows on the building (the force application point) nor which direction the wind is blowing, because the building is stationary ($\mathbf{d} = \mathbf{0}$).
You can similarly construct more examples based on my approach. I hope that helps.
