Question

A force f= 1i + 4j acts on the block shown .the fore of friction acting on the block is

Answer 1

Answer: The force of friction acting on the block is 2.94 N.

Explanation:

Given that,

Mass m = 1kg

Coefficient of friction $\mu = 0.3$

We know that,

The frictional force is the product of the coefficient of friction and normal force.

$F_{\mu} =\mu N$

Where, N = normal force

Normal force is the product of the mass and gravitational acceleration.

N = mg

Now,

The frictional force is

$F_{\mu} = \mu\times mg$

$F_{\mu} = 0.3\times 1\ kg\times 9.8\ m/s^{2}$

$F_{\mu} = 2.94 N$

Hence, the force of friction acting on the block is 2.94 N.

Answer 2

"Given, Mass m = 1Kg;

Coefficient of friction

$\mu \quad =\quad 0\cdot 3;$

By definition, $F\mu \quad =\quad \mu N$

Where,$F\mu$= Frictional force; N= normal force

We know that N=mg

Where normal force is the product of mass and gravitational acceleration

The frictional force is, $F\mu \quad =\quad \mu \times mg$

$F\mu \quad =\quad 0\cdot 3\quad \times \quad 1Kg\quad \times \quad \frac { 9\cdot 8m }{ { s }^{ 2 } }$

$F\mu \quad =\quad 2\cdot 94N$

Therefore, the force of friction,

$F\mu \quad acting \quad on \quad block \quad is \quad 2.94N$ "