Question

a block of mass 5 kg is placed on a rough horizontal surface having new is equal to 0.4 a force of 13 newton is applied in horizontal direction calculate frictional force acting on block and its acceleration

Answer 1

m= 5kg
U(coefficient of friction)= 0.4
F= 13N
N=mg= 50N
f(Friction)=0.4×50= 20N
As limiting Friction is greater than Force body does not move so
Acceleration=0
Hope your doubt is cleared

Answer 2

A block of mass 5kg is placed on rough horizontal surface having $\mu$ is equal to 0.4. and a force of 13N is applied in horizontal direction.

Here, F = 13N
coefficient of friction,$\mu$ =0.4
we know, $F_f=\mu N$
where N is normal reaction act on block by rough surface. e.g., N = weight of body = mg
e.g., $F_f=0.4\times5\times10$
$F_f=20N$

here we can see that frictional force is greater than applied force. so, body can't move .
hence, acceleration of body = 0.