Question

Question :-
A block of mass 2 kg is placed on the floor. As shown in diagram. Find the value of friction force between the block and the floor ?

Answer 1

Given :-

• Mass of block, m = 2kg
• Force applied on block, F = 2.5 N
• the coefficient of static friction , μ = 0.4
• Acceleration of gravity, g = 10 m s-2

Info :-

• the ratio of friction force between two bodies in which the force pressing them together is the coefficient of static friction, which is denoted by μ.

Solve :-

• Maximum static frictional force,

• fmax = µs × m × g = 0.4 × 2 × 9.8

• fmax = 7.84 N

Since,

~ the applied force on the block is less than the maximum static frictional force and the frictional force on the block is equal to the applied force = 2.5 N.

And this Happened because static friction is a self adjusting force

${}$

Answer 2

Question :-

A block of mass 2 kg is placed on the floor. As shown in diagram. Find the value of friction force between the block and the floor ?

Given :-

• Mass of block, m = 2kg

• Force applied on block (F) = 2.5N

• Coefficient of static friction, µ = 0.4

• Acceleration of gravity = 10m/s²

To find :-

Value of friction between the block and the floor.

Solution :-

The maximum static force is µmg = 0.4 × 2 × 10 = 8N

The value of friction force between the block and the floor = 2.5N

Explanation :-

8N is more than applied force. When a force of 2.5N is applied, static frictional force is also 2.5N as it is self adjusting force.

The friction adjusts its value and a new value becomes 2.5N so that block remains equillibrium.