A 2.5 kg block initially at rest is pulled to the right along a horizontal, frictionless surface
by a constant, horizontal force of 12.3 N.
Find the speed of the block after it has
moved 2.9 m.

The answer is a 0.64 volts

W= fd
where f is an applied force and d is the distance over which it is applied. when a parachuting skydiver is at constant velocity, he is experiencing "terminal velocity", the point at which his acceleration down is canceled out by air resistance. that means that the resisting force of the air is equal to his weight, but in the opposite direction. with this, we can solve for the work done by the force.
w = fd
w = (92.0)(9.81 m/s^2) (325)
w= 2.93 x 10^5 j
the value for work in this case is positive because it is pushing up on the skydiver, who has a downward force of gravity.
therefore the answer is 2.93 x 10^5 j

You need to be more specific.  i really don't know what a "motion map" is.you can draw a graph of an object's position, distance, displacement, speed,velocity, or acceleration versus time, or you can draw a graph of the object's x-position versus its y-position and not give any information about time at all.if the graph presents position or displacement against time,  then constant velocity is a straight line . . i.e. a line that has  constant slope.if the graph shows speed against time, then constant velocityis a horizontal piece of line.if the graph shows acceleration against time, then constant  velocity is a piece of line that lies on the x-axis . . it's value is zero.