A particle of mass 0.1 kg is subjected to a force which varies with distance as shown in the figure. If it starts its journey from rest at x = 0, its velocity at x = 12 m is
40 m/s
Work done = Area under the F-x graph
=12×10×4+10×4+12×10×4=80 J
Now, work done = increase in kinetic energy. If v is the velocity at x = 12 m, then increase in K.E. =12mv2.
∴12mv2=80
⇒v2=80×2m=80×20.1=1600
⇒v=40 ms−1
Hence, the correct choice is (d).