Question

sam of mass 40 kg is slipping on the frost.if the coefficient of friction acting is 0.45 find the frictional force acting between him and frost layer​

Answer 1

Given :-

Mass of Sam = 40 kg

Coefficient of friction = 0.45

To Find :-

The frictional force acting between him and frost layer.

Analysis :-

Here we are given with the mass and the coefficient of friction.

Firstly find the normal force by substituting the given values in the question accordingly.

Then using the friction formula, find the frictional force accordingly.

Solution :-

We know that,

• m = Mass
• μ = Coefficient of friction
• g = Gravity
• n = Normal force

Using the formula,

$\underline{\boxed{\sf Normal \ force=Mass \times Gravity}}$

Given that,

Mass (m) = 40 kg

Gravity (g) = 9.8 m/s

Substituting their values,

⇒ 40 × 9.8

⇒ 392 N

Using the formula,

$\underline{\boxed{\sf Frictional \ force=\mu N}}$

Given that,

Coefficient of friction (μ) = 0.45

Normal force (n) = 392

Substituting their values,

⇒ 0.45 × 392

⇒ 176.4 N

Therefore, the frictional force acting between him and frost layer​ is 176.4 N.

Answer 2

$\Large\bf{\color{coral}GiVeN,}$

• Mass of Sam (m) = 40 kg.

• Coefficient of friction between Sam and frost $\bf{(\mu)}$ = 0.45

$\Large\bf{\color{lime}To\:FiNd,}$

• The frictional force acting between Sam and frost layer.

$\Large\bf{\color{cyan}CaLcUlAtIoN,}$

✒ See the attachment picture.

$\bf\blue{We\:have}$

$\red\bigstar\:\:\bf{\color{peru}Normal\:Force\:(F_N)\:=\:m\:g\:}$

$\bf\pink{Where,}$

• m = 40 kg

• g = $\bf{9.8\:m/s^2}$

$:\implies\:\:\bf{Normal\:Force\:(F_N)\:=\:40\times{9.8}\:}$

$:\implies\:\:\bf\orange{Normal\:Force\:(F_N)\:=\:392\:N\:}$

$\bf\purple{Now,}$

☆ The frictional force is articulated by,

$\pink\bigstar\:\:\bf{\color{olive}Frictional\:Force\:(F_f)\:=\:\mu\:F_N\:}$

$:\implies\:\:\bf{Frictional\:Force\:(F_f)\:=\:0.45\times{392}\:}$

$:\implies\:\:\bf\green{Frictional\:Force\:(F_f)\:=\:176.4\:N\:}$

$\Large\bold\therefore$ The frictional force acting between Sam and frost layer is 176.4 N.