A block of mass 2 kg is placed on the floor. The coefficient of static friction is 0.4. A horizontal force of 2.5 N is applied on the block. Calculate the force of friction between the block and the floor.

2.5N

Explanation:

Given the following :

Mass of block (m) = 2kg

Coefficient of static friction (μs) = 0.4

Horizontal force applied to the block = 2.5N

The frictional force (Ff) between the block and the floor is :

First calculate the maximum static frictional force:

Frictional force = Coefficient of static friction(μs) × normal reaction(R)

Normal reaction(R) = mass × acceleration due to gravity (10m/s^2)

R = 2 × 10 = 20

Fmax = μs × R

Fmax = 0.4 x 20 = 8N

Here, since the applied force (2.5N) is less than maximum frictional force(8N).

The force of friction between the block and the floor will be equal to the applied force of 2.5N due its ability to adjust itself in other to ensure equilibrium.

volts times amps 6 x 0.5 3w

its c i just checked it