In the figure shown masses of the blocks A, B and C are 6kg, 2kg and 1kg respectively. Mass of the spring is negligibly small and its stiffness is 1000 N//m. The coefficient of friction between the block A and the table is `mu =0.8`. Initially block C is held such that spring is in relaxed position. The block is released from rest. Find `(g=10 m//s^(2))`.

(a) the maximum distance moved by the block `C`.
(b) the acceleration of each block, when elongation in the spring is maximum.

Explanation

Block `A` and `B` will not move unless tension in the string `T ge mu_(A)g(48 N)`.
Let blocks `A` and `B` do not move and maximum elongation in the spring be `Deltax`.
Applying conservation of mechanical energy for the block `C` and the spring. We get
`Deltax=(2m_(c )g)/(k)=2xx10^(-2)m`
In this situation F.B.D. of the block `B` is
For blocks to be in equilibrium `T=kx_(0)+mg=40N`.
`:' T_(max)=40N( lt 48N)` Hence blocks `A`
Therefore maximum distance moved by the
block `C` is `Deltax_(0)=(2m_(c )g)/(k)=2xx10^(-2)m=2cm`