If force, time and velocity are treated as fundamental quantities then dimensional formula of energy will be

Explanation

Let the quantity be Q, then
`Q=f (v, F, t)`
Assuming that the function is the product of power functions of v, F and T,
`Q=kv^(x)F^(y)T^(z)` …(i)
where K is a dimensionless constant of proportionality. The above equation dimensionally becomes
`[Q]=[LT^(-1)]^(x)[MLT^(-2)]^(y)[T]^(z)`
i.e., `[Q]=[M^(y)][L^(x+y)T^(-x-2 y+z)]` ...(ii)
Now
(a) Q= mass i.e., `[Q]=[M]`
So Eqn. (ii) becomes
`[M]=[M^(y)L^(x+y)T^(-x-2 y+z)]`
its dimensional correctness requires
`y-1," "x+y=0` and `-x-2y+z=0`
which on solving yields
`x=-1," "y=1` and `z=1`
substituting it in Eqn. (i) we get
`Q=kv^(-1) FT`
(b) Q = Energy
t.e., `[Q]=[ML^(2)T^(-2)]`
So, Eqn. (ii) becomes
`[ML^(2)T^(-2)]=[M^(y)L^(x+y)T^(-x-2 y+z)]`
which in the light of principle of homogeneity yields
`y=1, " "x+y=2` and `-x-2y+z=-2`
which on solving yields
`x = y = z = 1`
So, eqn. (i) becomes
`Q = kv FT`