Two block A and B of equal masses are placed on rough inclined plane as shown in figure. When and where will the two blocks come on the same line on the inclined plane if they are released simultaneously? Initially the block A is `sqrt2` m behind the block B. Co-efficient of kinetic friction for the blocks A and B are 0.2 and 0.3 respectively `(g=10m//s^2)`.

Answer

9.9

Explanation

We can find the accelerations of blocks by using
`a = (m g sin theta - mu_(2) m g cos theta)/(m)`
Acceleration of block A, `a_(A) = g sin theta - mu_(k,A) g cos theta = 4 sqrt(2) m//s^(2)`
Acceleration of blockB, `a_(B) = g sin theta - mu_(k,B) g cos theta = 3.5 sqrt(2) m//s^(2)`
`a_(AB)` is relative acceleration of A w.r.t. B, `a_(AB) = a_(A) - a_(B) = (1)/(sqrt(2)) m//s^(2)`
`L = sqrt(2) m, L = (1)/(2) a_(AB) t^(2) implies t = 2 s`
Distance moved by B during that time is given by
`S_(B) = (1)/(2) a_(B) t^(2) = (1)/(2) 3.5 sqrt(2) xx 4 = (2 xx 0.7)/(sqrt(2)) xx 10 = 7 sqrt(2) m`
Similarly for A, `S_(A) = 8 sqrt(2) m`