A block A of mass 2 kg is connected with two springs, as shown . The spring constant of lower spring is system is thrice the spring costant of upper spring. The system is released from rest with both the springs unstreched. The maximum displacement of bloack is `0.1 m`. Find the acceleration of the block is its lowest position-

Options

A. `6.5 m//s^(2)`
B. `7.5 m//s^(2)`
C. `10 m//s^(2)`
D. `8 m//s^(2)`

Answer

C

Explanation

`mg x = 1/2 (k + 3k) x^(2)`
`x = (10)/(k) = 0.1` so `k = 100 N//m`
`A = ((k + 3k)x - mg)/(m) = 10 m//s^(2)`