Question

Two blocks a and b of equal masses are sliding down along straight line on an incline plane of 45° coefficient of kinetic friction arm you equal to 0.2 and b equal to 0.3 respectively. at equal to zero both the blocks are at rest and block a is root 2 metre behind block b that time and distance from the initial position where the front face of the blocks come in the line on the incline plane as

Answer 1

acceleration of block a is given by

$a_1 = \frac{mgsin\theta - \mu mgcos\theta}{m}$

$a_1 = g(sin45 - 0.2*cos45) = 5.54 m/s^2$

similarly for another block

$a_2 = \frac{mgsin\theta - \mu mgcos\theta}{m}$

$a_2 = g(sin45 - 0.3cos45) = 4.85 m/s^2$

now the distance between two blocks initially is given as

$d = \sqrt2 m$

now we can use kinematics to find the time after which two blocks are in same line

$d = v_i * t + \frac{1}{2} at^2$

here vi = 0

and here acceleration must be relative acceleration of two blocks

plug in all values

$\sqrt2 = 0 + \frac{1}{2}(5.54 - 4.85)*t^2$

$t = 2 s$

so two blocks will be in same line after 2 s

now distance traveled by first block in the same time

$d = v_i*t + \frac{1}{2} at^2$

$d = 0 + \frac{1}{2}*5.85*2^2$

$d = 11.7 m$

so block will move a total distance of 11.7 m till that time