If a rod of length  , very small area of cross-section A, Young’s modulus of elasticity Y is acted upon by two parallel forces 3F and F respectively (as shown) and placed on a smooth horizontal plane. If the elastic limit is not crossed and then to study the change in length of rod Δl and it’s elastic potential energy (U) the rod is segmented into four equal parts where magnitude of change in lengths are Δl1,Δl2,Δl3,Δl4 and elastic potential energy stored in each segment are U1,U2,U3,U4 respectively as shown then which is/are correct?

In steady state, the F.B.D. of the rod can be drawn by dividing it into four  equal parts as shown. We have  YΔll=stressaverage⇒Δl=lY.FavgA=lY.F1+F22A  For any rod . If   Now,  Δl1compression=Favg(length1)AY=F+02l4AY=Fl8AY;Δl2elongation=Fl8AY acceleration of the system=a=3F+F4m=FmFrom FBD of all blocks we get N1,N2 and N3 are the normal reactions betweeen 1st and 2nd blocks,  2nd  and  3rd blocks, and  3rd  and  4th blocks on solving we get N1=0, N2=F and N3=2FΔl3( elongation )=2F+F2l4AY=3Fl8AY,Δl4( elongation )=2F+3F2l4AY=5Fl8AYΔl=8FI8AY=FlAY ΔU∝Strain 2⇒ΔU∝(Δl)2