Question

A force of 20N acts towards the right on a block of 2kg on a rough surface. the coefficient of static friction for the 2 surfaces is 0.6 and the coefficient of Kinetic friction is 0.4. Find the friction force acting on the block and the acceleration of the block.

Answer 1

Answer:

Explanation: 1st find static friction , 0.6 x 20 = 12 which is less than the

                ext force i.e, 20 N

            now the body will move so friction shifts its value to kinetic

                then ,      20 - 0.4 x 20 = 2 a

                          then a = 6 m/s²

hope u get this n let me know...................................  

Answer 2

Solution:-The block will not move untill the horizontal force is not much more than the,static friction.....

so,here

max.static friction=coefficient of static friction×N

=0.6×2×10

=12N

here the horizontal force is 20N,which means it is much more than then static friction,,,so it will move

now,when the object starts to move,,then kinetic Friction acting on it=0.4×20=8N

now here

resultant force=20-8=12

acceleration=resultant force/mass=12/2=6

hence, acceleration will be 6m/sec^2