Write the relationship between-
(i) kgf and newton
(ii) gf and dyne

Hint:All the given units are units of force. Kgf is the gravitational metric unit, and Newton is S.I. unit of force, gf is a metric unit of force, and dyne is the unit of force in a centimeter-gram-second system. We will be establishing the relationship between all the given units by converting them into the S.I. unit system.

Complete step by step answer:
(i) One kilogram-force (kgf) is the value of force due to gravity on a body of mass \[1{\rm{ kg}}\]. Mathematically, we can write:
\[1{\rm{ kgf}} = mg\]……(1)
Here m is the mass of the body, and g is the acceleration due to gravity.
We know that the value of acceleration due to gravity is \[9.81{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}\].
Substitute \[9.81{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}\] for g and \[1{\rm{ kg}}\] for m in equation (1).
\[
1{\rm{ kgf}} = \left( {1{\rm{ kg}}} \right)\left( {9.81{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}} \right)\\
\Rightarrow 1{\rm{ kgf}} = 9.81{{{\rm{ kg}} \cdot {\rm{m}}} {\left/
 {\vphantom {{{\rm{ kg}} \cdot {\rm{m}}} {{{\rm{s}}^2} \times \left( {\dfrac{{\rm{N}}}{{{\rm{kg}} \cdot {{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}}}} \right)}}} \right.
} {{{\rm{s}}^2} \times \left( {\dfrac{{\rm{N}}}{{{\rm{kg}} \cdot {{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}}}} \right)}}\\
\Rightarrow 1{\rm{ kgf}} = 9.81{\rm{ N}}
\]
(ii) Gram force (gf) is the value of force due to gravity on a body of mass \[1{\rm{ g}}\].
\[1{\rm{ gf}} = {m_1}g\]
Here \[{m_1}\] is the mass of the body.
Substitute \[9.81{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}\] for g and \[1{\rm{ g}}\] for \[{m_1}\] in the above expression.
\[
1{\rm{ gf}} = \left( {1{\rm{ g}}} \right)\left( {9.81{\rm{ }}{{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}} \right)\\
\Rightarrow 1{\rm{ gf}} = 9.81{\rm{ g}} \cdot {{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}} \times \left( {\dfrac{{{\rm{kg}}}}{{1000{\rm{ g}}}}} \right)\\
\Rightarrow 1{\rm{ gf}} = 9.81 \times {10^{ - 3}}{\rm{kg}} \cdot {{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}} \times \left( {\dfrac{{\rm{N}}}{{{\rm{kg}} \cdot {{\rm{m}} {\left/
 {\vphantom {{\rm{m}} {{{\rm{s}}^2}}}} \right.
} {{{\rm{s}}^2}}}}}} \right)\\
\Rightarrow 1{\rm{ gf}} = 981 \times {10^{ - 5}}{\rm{ N}}\]……(2)
We know that the value of dyne in terms of newton is given as:
\[1{\rm{ dyne}} = {10^{ - 5}}{\rm{ N}}\]
Substitute \[1{\rm{ dyn}}\] for \[{10^{ - 5}}{\rm{ N}}\] in equation (2).
\[
1{\rm{ gf}} = 981 \times \left( {1{\rm{ dyn}}} \right)\\
\therefore 1{\rm{ gf}} = 981{\rm{ dyn}}
\]

Therefore, the relation between one kgf is equal to \[9.81{\rm{ N}}\] , and one gf is equal to \[981{\rm{ dyn}}\].

Note: It would be better if we remember converting units of force into different units of units.There are various systems of units such as the centimetre-gram-second (CGS) system, foot-pound-second (FPS) system, and system Internationale (S.I.) system. But the S.I. the unit system is mostly followed by mathematicians and scientists.