1) A body constrained to move along the z-axis of a coordinate system is subjected to a constant force given by, F = - i cap + j cap +3 k cap N, where i cap, j cap, k cap are unit vectors along the x, y, z- axis of the system, respectively. Which is the work done by this force in moving the body a distance of 4 m along the z -axis ? (answer should come 12 Joules)

2) A body is thrown on a rough surface such that friction force acting on it is linearly varying with distance travelled by it as f = ax + b. Find the work done by the friction on the box if before coming to rest the box travels a distance s. (answer should come  \frac{1}{2} a s^{2} + bs)

3) You are re-shelving books in library. You lift a book from the floor to the top shelf. This kinetic energy of the book on the floor was zero and the kinetic energy of the book on the to shelf is zero, so no change occurs in the kinetic energy yet you did some work in lifting the book. Is the work-kinetic energy theorem violated? (answer should come -- there is no violation, total work on the book is zero.)




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Vector  force  F = (-1 * i + 1 * j + 3 k ) Newtons
   Displacement of the body = vector s =  4 k  meters

work done =  F . s = dot product of force and displacement
           = - 1*4 (i . k)  + 1*4 (j . k ) + 3*4 ( k . k)  joules
             = 0 + 0 + 3*4 = 12 joules
Work is done only along z direction by the z component of force.  So multiply these two.
  Friction force  =  F = a x + b ,  where x is distance travelled.
  work done in moving the body by a small distance dx with a force F = (ax+b) at point x away from the initial starting point, is  F . dx = (ax +b ) dx.

Summation of the tiny pieces of work done in small distances dx, moving from x = 0 to x = s, is found by integration.  Integration is compulsory, as F varies wrt x.

 \int\limits^s_0 {(ax+b)} \, dx = [ ax^2/2 + b x ]_0^s = \frac{1}{2}as^2+b s\\


It is a tricky question..

Here, in moving the book from the floor to the top shelf, there is a change in the potential energy.   The work done is equal to the change in potential energy.  PE change = m g h.

If the potential energy remained same, ie., the book is at the same height, we could say, that there is no work done on the book.  Not in this case.
However, Let us look at it in this way.   we are talking about the work done in moving the book from floor and then moving it with a uniform velocity and then placing it at rest, -- only these three steps. 

Let the book have initial kinetic energy 0. Let the hands give a kinetic energy KE1 so it starts moving at uniform velocity.  Then after reaching the top shelf, it has been slowed down to be at rest again.  So the kinetic energy KE1 is lost or absorbed by the hands.  Hence, the net change in KE is 0.  Net work done is = net change in KE = (KE1-0) + (0-KE1) = 0

But taking a bigger picture combining with the potential energy change in Earth's gravitational field, we say there is work done.

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