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.
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.