Two blocks A and B each having a mass of 20kg, rest on frictionless surfaces as shown in the figure below.Assuming the puelleys to be light and frictionless, compute `:`
(a) the time required for block A, to move down by 2m on the plane, starting from rest,
( b) tension in the string, connecting the blocks.

Explanation

Step 1. Draw the FBDs for both the blocks. If tension in the string is T, then we have

Note that `m_(A) g `, shown better be resolved along and perpendicular to the plane, as the block A is moving along the plane.

Step 2. From FBDs, we write the force equations 'for block A where
`N_(A) = m_(A) g cos theta = 20 xx 10 xx 0.8 = 160 N ` and `m_(A) g sin theta - T = m_(A) a ` ...(i)
Where 'a' is acceleration of masses of blocks A and B.
Similarly, force equations for block B are `N_(B) = m_(B) g = 20 xx 10 = 200 N `
and `T = m_(B ) a ` ....(ii)
From (i) and (ii), we obtain
`a = ( m_(A) gsin theta )/( m_(A) + m_(B)) = ( 20 xx 10xx 0.6 ) /( 40) = 3 ms^(-2)`
`T = m_(A) a = 20xx 3 = 60 N `
Step 3. With constant acceleration `a = 3ms^(-2)`, the block A moves down the inclined plane a distance S = 2m in time t given by
`S - ( 1)/( 2) at^(2) ` or `t = sqrt((2S)/( a)) = ( 2)/( sqrt(3)) ` seconds