Question

Figure shows $N$ identical blocks connected with identical springs on a smooth horizontal surface. A constant force is pulling the blocks horizontally. During motion all the springs have acquired their stable configurations. If stretch in spring connecting 4th and 5th block is two times that between 8th and 9th block what is total number of blocks

• A. $6$
• B. $8$
• C. $10$
• D. $12$

Answer 1

Answer: (D)

$12$

force pulling all the$N$ blockes $=P$

Let the acceleration of all blocks is $a$.

so $N.ma=P$

now let spring constant is given by $K$.

then force required to pull ($N-1$) blocks (removing 1st block) is $(N-1).ma$ and that will be equal to $K.x_1$, where $x_1$ is stretch in spring between block 1 and 2.

similarly $(N-2).ma$ for block 2 and 3

and so on .$(N-4).ma$ for blocks 4 and 5 , and $(N-8).ma$for block 8 and 9.

according to the condition given in the question $(N-4).ma=2.(N-8).ma$

after solving we get $N=12$.