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Question: If the velocity (V), force (F), and time (T) are chosen as fundamental quantities, Express : a) Ma...

If the velocity (V), force (F), and time (T) are chosen as fundamental quantities, Express :
a) Mass
b) Energy, in terms of V, F, and T.

Explanation

Solution

Write the dimensional formula of velocity (V), force (F), and time (T), then express the required quantity in new fundamental quantities with assumed powers and compare the dimensional powers of both of them to calculate the value of assumed coefficients.

Complete step by step solution:
(a) We know that dimensions of velocity (V), force (F), and time (T) are [M0L1T−1],[M1L1T−2],[T1]\left[ {{M^0}{L^1}{T^{^{ - 1}}}} \right],\left[ {{M^1}{L^1}{T^{ - 2}}} \right],\left[ {{T^1}} \right] respectively. Assume
Mass=(constant)[FbVaTc]Mass = (constant)\left[ {{F^b}{V^a}{T^c}} \right]
Now by equating the dimensions we get:-
[M1]=[M0L1T−1]a[M1L1T−2]b[T]c\left[ {{M^1}} \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^a}{\left[ {{M^1}{L^1}{T^{ - 2}}} \right]^b}{\left[ T \right]^c}
⇒[M1]=[MbLa+bTc−2b−a]\Rightarrow \left[ {{M^1}} \right] = {\left[ {{M^b}{L^{a + b}}{T^{c - 2b - a}}} \right]^{}}
So by equating the powers we get
b=1b = 1 --(i)
a+b=0a + b = 0 --(ii)
c−a−2b=0c - a - 2b = 0 --(iii)
Now by substituting the value of b in equation (i) we get
a+1=0 ⇒a=−1  a + 1 = 0 \\\ \Rightarrow a = - 1 \\\
Now by substituting the values of a and b in equation (iii)
c+1−2=0 ⇒c=1  c + 1 - 2 = 0 \\\ \Rightarrow c = 1 \\\
Hence, We can write dimensions of mass as Mass=(constant)[V−1F1T1]Mass = (constant)\left[ {{V^{ - 1}}{F^1}{T^1}} \right]

(b)Now we will write the energy in terms of velocity(V), force(F), and time(T). We know that the dimensions of energy are[ML2T−2]\left[ {M{L^2}{T^{ - 2}}} \right].
Assume
Energy=(constant)[VaFbTc]Energy = (constant)[{V^a}{F^b}{T^c}]
Now by equating the dimensions we get
[ML2T−2]=[M0L1T−1]a[MLT−2]b[T]c\left[ {M{L^2}{T^{ - 2}}} \right] = {\left[ {{M^0}{L^1}{T^{ - 1}}} \right]^a}{\left[ {ML{T^{ - 2}}} \right]^b}{\left[ T \right]^c}
⇒[ML2T−2]=[MbLa+bTc−2b−a]\Rightarrow \left[ {M{L^2}{T^{ - 2}}} \right] = \left[ {{M^b}{L^{a + b}}{T^{c - 2b - a}}} \right]

Now by equating the powers of the mentioned quantities. We will get
b=1b = 1 --(i)
a+b=2a + b = 2 ---(ii)
c−2b−a=−2c - 2b - a = - 2 ---(iii)
So, by putting the value of b in equation (ii) we will get
a=1a = 1
Similarly, by putting the value of a and b in equation (iii) we will get
c−2−1=−2 ⇒c=1  c - 2 - 1 = - 2 \\\ \Rightarrow c = 1 \\\
Hence, We can write dimensions of energy as Energy=(constant)[V1F1T1]Energy = (constant)[{V^1}{F^{^1}}{T^1}]

Note: When solving for assumed coefficients, first solve the equation which has only one variable involved then solve the subsequent equations by substituting the values of already found variables, this minimizes the chances of error in calculation.