
Explain the meaning of the following statement: $1kgf = 9.8N$
Answer
491.7k+ views
Hint: Both are units of force, as it is quite evident in the question. A push or pull acting on an object is called the force acting on the object. It affects the motion of the object and causes some acceleration or deceleration of the object. It is also dependent on the mass of the object.
Complete step by step answer:
$1{\text{ }}Kgf$ is the force with which Earth pulls a body of mass $1{\text{ Kg}}$ towards itself.
The gravitational unit of force in the M.K.S system is kilogram-force ( $Kgf$).
Force= mass$ \times $ acceleration
And the S.I. The unit of force is Newton.
Therefore,
$1{\text{ }}Kgf$ = gravitational force on a mass of $1{\text{ }}Kg$
$\Rightarrow 1{\text{ }}kg$ mass $\times$ acceleration due to gravity ($g$ )
$\Rightarrow g$ Newton
Here, $g = 9.8m{s^{ - 2}}$
Acceleration due to gravity is the acceleration of the body during free fall. It's S.I. unit is $m{s^{ - 2}}$. It is a vector quantity since it has both magnitude and direction. It is based on Newton’s second law of motion, which says force experienced by a body is the mass times acceleration applied to it.
Near the Earth’s surface, acceleration due to gravity has an approximately constant value, which is $9.8{\text{ }}m{s^{ - 2}}$.
$g = \dfrac{{GM}}{{{r^2}}}$
Where $G$ is the universal gravitational constant. $6.673 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
$M$ is the mass of Earth. $M = 6 \times {10^{24}}kg$
$r$ is the distance of the center of the mass of the body from the large attracting body.
$r$ = radius of earth = $6.4 \times {10^6}m$
$g = \dfrac{{GM}}{{{r^2}}}$
$\Rightarrow g = \dfrac{{6.673 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{{{(6.4 \times {{10}^6})}^2}}}$
$\Rightarrow g = 9.8m{s^{ - 2}}$
$g$ is dependent on the distance of the center of mass of the body from the large attracting body, i.e., the radius of the Earth. Hence its value fluctuates on the poles and equator because Earth is in the shape of an ellipsoid.
Note: Every entity has its own International System of Units (S.I. unit). Besides, there exists the M.K.S and F.P.S systems. M.K.S stands for meter, kilogram, and second as base units. F.P.S stands for foot, pound, and second as base units.
Complete step by step answer:
$1{\text{ }}Kgf$ is the force with which Earth pulls a body of mass $1{\text{ Kg}}$ towards itself.
The gravitational unit of force in the M.K.S system is kilogram-force ( $Kgf$).
Force= mass$ \times $ acceleration
And the S.I. The unit of force is Newton.
Therefore,
$1{\text{ }}Kgf$ = gravitational force on a mass of $1{\text{ }}Kg$
$\Rightarrow 1{\text{ }}kg$ mass $\times$ acceleration due to gravity ($g$ )
$\Rightarrow g$ Newton
Here, $g = 9.8m{s^{ - 2}}$
Acceleration due to gravity is the acceleration of the body during free fall. It's S.I. unit is $m{s^{ - 2}}$. It is a vector quantity since it has both magnitude and direction. It is based on Newton’s second law of motion, which says force experienced by a body is the mass times acceleration applied to it.
Near the Earth’s surface, acceleration due to gravity has an approximately constant value, which is $9.8{\text{ }}m{s^{ - 2}}$.
$g = \dfrac{{GM}}{{{r^2}}}$
Where $G$ is the universal gravitational constant. $6.673 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
$M$ is the mass of Earth. $M = 6 \times {10^{24}}kg$
$r$ is the distance of the center of the mass of the body from the large attracting body.
$r$ = radius of earth = $6.4 \times {10^6}m$
$g = \dfrac{{GM}}{{{r^2}}}$
$\Rightarrow g = \dfrac{{6.673 \times {{10}^{ - 11}} \times 6 \times {{10}^{24}}}}{{{{(6.4 \times {{10}^6})}^2}}}$
$\Rightarrow g = 9.8m{s^{ - 2}}$
$g$ is dependent on the distance of the center of mass of the body from the large attracting body, i.e., the radius of the Earth. Hence its value fluctuates on the poles and equator because Earth is in the shape of an ellipsoid.
Note: Every entity has its own International System of Units (S.I. unit). Besides, there exists the M.K.S and F.P.S systems. M.K.S stands for meter, kilogram, and second as base units. F.P.S stands for foot, pound, and second as base units.
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