Question

If velocity (V), force (F), and time (T) are chosen as fundamental quantities, express
1.Mass and
2.Energy
In terms of V, F and T

Answer 1

Hint: The given question belongs to the chapter Units and dimensions so, first let us see the definition of dimension.
DIMENSION: dimensions are the powers to which fundamental quantities are raised to express the given physical quantity.
Concept of dimensions is used to find the relation between different physical quantities if the factors on which they depend are known.

Complete step by step solution
We have to express mass in terms of velocity, force and time so, let they are related as
\[M = {V^a}{F^b}{T^c}\] now we know that if a given equation is correct then the dimension of both side should be same on comparing the dimension of both side of the equation we get\[{M^1}{L^0}{T^0} = {M^b}{L^{a + b}}{T^{ - a - 2b + c}}\] from the equation we get \[a = - 1,b = 1,c = 1\] and hence the mass can be expressed as \[M = {V^{ - 1}}{F^1}{T^1}\]
Here we have to express energy in terms of velocity (V), force (F), and time (T) so, let energy is related to V, F, and T as \[E = {V^a}{F^b}{T^c}\] now on comparing the dimension we the equation\[{M^1}{L^2}{T^{ - 2}} = {M^b}{L^{a + b}}{T^{a - 2b + c}}\] on comparing the dimension we get\[a = 1,b = 1,c = - 1\] and Hence the energy can be expressed in terms of V, F, and T as \[E = {V^1}{F^1}{T^{ - 1}}\]


Note: Mass and energy are actually related to velocity (V), force (F), and time (T) as \[M \propto {V^a}{F^b}{T^c}\] and \[E \propto {V^a}{F^b}{T^c}\] but above we are directly equating because on replacing the proportionality sign we have to introduce a constant which will be dimensionless so it will affect the dimensional formula of the above quantities.