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))`.

Answer

A, B, C

Explanation

`mum_(A)g = 0.8 xx 6 xx 10 = 48 N`
`(m_(B) + m_(C))g = (1+2) xx 10 = 30 N`
Since `(m_(B) - m_(C))g gt mum_(A)g, a_(A) = a_(B) = 0`.
From conservation of energy principal we can prove that maximum distance moved by `C` or maximum extension in the spring would be
`x_(m) = (2m_(C)g)/(k) = (2 xx 1 xx 10)/(1000)`
`= 0.02 m`

At maximum extension
`a_(C) = (kx_(m) - m_(C)g)/(m_(C))`
Substituting the values we have,
`a_(C) = 10 m//s^(2)`.