Without calculus

the net force on the system is 30N, now let us take the mass of the rod to be 'm', Then its acceleration is $\frac { 30 }{ m }$

Now let us switch to that frame, and since that frame is non inertial the rod experiences a pseudoforce of $\frac { 30 }{ m } .m$

now , the situation is as shown below ( in C frame )

Situation

In this frame rod is in equillibrium,, also as pseudoforce acts on all particles ,we can consider it equivalent to be passing through com,

Thus considering the equillibrium of each half of rod seperately, we

see the situation to be identical to two rods each having half the length one experiencing a force of 10 N on either side thus under a tension of 10 N and other 40 N

Thus seperately finding elongation

$\frac { { f }_{ 1 }l }{ 2AY } +\frac { { f }_{ 2 }l }{ 2AY } =\Delta ({ l }_{ 1 }+{ l }_{ 2 })=\Delta l$

which is 1m,

thus an s 100

NOTE

1) ignore the arrow marks on individual halves

2) The common calculus based method involves considering a thin element of rod and finding tension as a function of x, and thus calculating the small elementary elongation of each component and adding it up

3) Now you can remember the result as the situation equivalent to having T=average of scalar sum of forces

4) I think my method is simpler

Tension decreases linearly along the rod as we move from right to left. Therefore, instead of integrating for each small element of rod, we can directly take the average i.e. we can assume uniform tension of (10+40)/2=25 N along the rod. Then using Hooke's law, we directly get l=1m.

Ans: 100

Relevant wiki: elastic

Tension decreases linearly along the rod as we move from right to left.

Therefore, instead of integrating for each small element of rod, we can directly take the average

i.e. we can assume uniform tension of (10+40)/2=25 N along the rod.

Then using Hooke's law, we directly get l=1m.