Question

See the attachment for question 30 please answer fast with solution and answer is A

Answer 1

Given:

Horizontal force = 6 N

Vertical applied force = 9 N

Coefficient of static Friction = 0.4

To find:

Frictional force ?

Concept:

We first need to find the max frictional force after which the block starts moving , its called the Limiting Condition .

Calculation:

Let limiting friction be f

$\therefore \: f = \mu \times N$

$= > f = 0.4 \times \bigg \{(2.5 \times 10) - 9 \bigg \}$

$= > f = 0.4 \times \bigg \{25 - 9 \bigg \}$

$= > f = 0.4 \times \bigg \{16 \bigg \}$

$= > f = 6.4 \: N$

Now applied force is 6.0 N , which is lesser than limiting friction .

Hence the block will not move and the static friction shall be equal to the applied force :

So friction force at that instant will be 6 N.

Answer 2

Static friction is used when the block is at rest and a force is acting on the block.

where as when the block starts to move, then we use kinetic friction.

!! Please note that coefficient of static friction and coefficient of kinetic friction may not be equal.

The solution is (A) 6.0N

Explanation is in the attachment below: