Two blocks A and B of equal masses are released from an inclined plane of inclination 45 ° at t = 0 . Both the blocks are initially at rest. The coefficient of kinetic friction between the block A and the inclined plane is 0 . 2 while it is 0 . 3 for block B . Initially the block A is 2 m behind the block B . At what time in seconds will their front faces come in a line, (Take g = 10 m s - 2 )

Step-by-step solution

Acceleration of A down the plane, a A = g sin 45° -μ A g cos 45 o = ( 10 ) ( 1 2 ) - ( 0.2 ) ( 10 ) ( 1 2 ) = 4√2m/s 2 Similarly acceleration of B down the plane, a B =g (sin 45°- μ B cos 45°) = ( 10 ) ( 1 2 ) - ( 0.3 ) ( 10 ) ( 1 2 ) = 3.5 2 m s 2 The front face of A and B will come in a line when, s a = s B + √2 or 1 2 u A t 2 = 1 2 a B t 2 + 2 1 2 × 4 2 × t 2 = 1 2 × 3.5 2 × t 2 + 2 Solving this equation, we get t = 2 s Additional information, s A = 1 2 a A t 2 = 1 2 × 4 2 × ( 2 ) 2 = 8 2 m So, both the blocks w