Question

A block of mass 2kg is placed on a rough surface, f=10n is applied on the block as shown force of friction of block is

Answer 1

,....mm..hfgjm.... hope u understand

Answer 2

Answer:

Coefficient of friction = $\mu =\ 0.5$

Explanation:

Given,

• Mass of the block =  m = 2 kg
• Force applied on the block = F = 10 N

We know that the friction force acting on the block due to the roughness of the surface and the block is equal to the multiplication of the coefficient of friction and the normal contact force between the surface and the block.

Let $\mu$ be the coefficient of the frictional force between the contact surface and the block.

$\therefore F\ =\ f\\\Rightarrow F\ =\ \mu mg\\\Rightarrow \mu\ =\ \dfrac{F}{mg}\\\Rightarrow \mu\ =\ \dfrac{10}{2\times 9.81}\\\Rightattow \mu\ =\ 0.50$

Friction force  $f\ =\ \mu mg\ =\ 0.5\times 2\times 9.81\ =\ 10\ N$