Question

The frictional force acting on 1 kg block is

$A.0.1N \\\ B.2N \\\ C.0.5N \\\ D.5N \\\$

Answer 1

Hint – Here we will proceed by using the formula of force (Newton’s second law) with S.I. unit kg into meter per second square. Then by applying the condition given in the diagram of 1 kg block we will get our answer.

Formula used, $F = M \times A$
Where, F = force
M = mass
A = acceleration

Complete step-by-step answer:
Friction Force – The friction force is a type of repulsive force due to rough surfaces between the blocks which restricts or opposes the motion of the object.

Total mass of the system = 1+100=101kg
Force acting on system(F) = 10N
Then, Acceleration of the system
$M \times a = F$ (M=101, F= 10)
We have to find
101$\times a$=10
$a = \dfrac{{10}}{{101}}$ meter per second$^2$
Maximum friction between two blocks is
$\mu mg = 0.5 \times 1 \times 10$ ( where g= 10 meter per sec$^2$)
$F\max = 5N$
But the external force that is $1 \times \dfrac{{10}}{{101}}$ is less than $F\max$.
Hence, both of them move together.
Therefore, the correction answer is option D.

Note – Whenever we come up with this type of question, one must know that Newton's second law of motion pertains to the behaviour of objects for which all existing forces are not balanced. The second law states that the acceleration of an object depends upon two variables – the mass of the object and the net force acting upon the object.