Question

Force of magnitude 4v2N is applied on block placed on rough horizontal
surface. If coefficient of friction between block and surface is 0.2, then
frictional force on the block will be
F=42N
45°
4 kg
u = 0.2
2N
3N
4N
5 N​

Answer 1

Given:

Force of magnitude 4√2 N is applied on block placed on rough horizontal surface. Coefficient of friction between block and surface is 0.2.

To find:

The friction experienced by the block ?

Calculation:

First of all, we need to calculate the normal reaction force between the block and the surface:

$\sf \therefore \: N + F \sin( {45}^{ \circ} ) = mg$

$\sf \implies \: N + \dfrac{F}{ \sqrt{2} } = mg$

$\sf \implies \: N + \dfrac{4 \sqrt{2} }{ \sqrt{2} } = 4 \times 10$

$\sf \implies \: N + 4 = 40$

$\sf \implies \: N = 36 \: Newton$

Now, the max static frictional force experienced by the block:

$\sf \therefore \: f = \mu_{s} \times N$

$\sf \implies\: f = 0.2 \times 36$

$\sf \implies\: f = 7.2 \: Newton$

But , forward force applied is :

$\sf \therefore \: F_{applied} = F \times \cos( {45}^{ \circ} )$

$\sf \implies\: F_{applied} = \dfrac{4 \sqrt{2} }{ \sqrt{2} }$

$\sf \implies\: F_{applied} = 4 \: Newton$

Since the applied force is lesser than the maximum frictional force , the static friction will regulate itself and will be equal to 4 N.

Hence frictional force is 4 N (i.e. block will be at rest).