A block of mass 2 ~kg is being pushed against a wall by a force F=90 ~N as shown in the figure. If the coefficient of friction is 0.25 , then the magnitude of acceleration of the block is (Take, .g=10 ~ms ⁻² ) ( 37^ = 3 5 )

Step-by-step solution

Block's weight (downward), w=2 10=20 ~N Vertical component of applied force (upwards), F_V=F =90 3 / 5=54 ~N Maximum frictional force, aligned F_r & = F =0 25 90 4 / 5 & =18 ~N aligned Net vertical force, F_ net =F_V- (F_r+w )=m a array ll & F_ net =54-18-20=16 ~N =m a & a= 16 2 =8 ~m / s ^2 array