Two blocks `A & B` attached to each other by a mass-less spring, are kept on a rough horizontal surface `mu=0.1`. A constant force `F=200 N` is applied on blcok `B` horizon tally as shown below. If a some instant the acceleration of `10 kg` mass is `12 m//s^(2)`, then the acceleration of `20 kg` mass is

Options

A. `2.5 m//s^(2) or 15.5 m//s^(2)`
B. `4 m//s^(2) or 10 m//s^(2)`
C. `3.6 m//s^(2) or 4.1 m//s^(2)`
D. `1.2 m//s^(2) or 1.3 m//s^(2)`

Answer

A

Explanation

(i) Let the spring is in a elongated state .

`f_(A) = 0.1 xx 10 g = 10 N` and
`f_(B) = -1 xx 20 g = 20 N `
`kx - 10 = 10 xx 12 implies kx = 130 N`
`200 - kx - 20 = 20 (a_(B)) implies a_(B) = 2.5 m//s^(2)`

(ii) Let the spring is in a compressed state .
`kx - 10 = 10 xx 12 impllies kx = 130 N`
`200 + kx - 20 = 20 (a_(B)) implies a_(B) = 15.5 m//s^(2)`