Answer to Question #170043 in Physics for gibson mkandawire

Question

The square of the speed v of an object undergoing acceleration a is a variable of a

function a and the displacement S according to the expression given by

𝑣² = 𝑘𝑎^𝑚𝑆^𝑛

What dimensions should k have in order for the expression to be dimensionally

consistent?

Answer 1

The units at the left hand side of the equality are:

$$[v^2] = \dfrac{m^2}{s^2}$$

The units at the right hand side of the equalily are:

$$[ka^mS^n] = [k]\dfrac{m^m}{s^{2m}}m^n = [k]\dfrac{m^{m+n}}{s^{2m}}$$

Comparing with the left hand side and expressing the dimension of $k$ , obtain:

$$\dfrac{m^2}{s^2} = [k]\dfrac{m^{m+n}}{s^{2m}}\\ [k] = \dfrac{m^2}{s^2}\dfrac{s^{2m}}{m^{m+n}} = m^{2-n-m}s^{2m-2}$$

where $m,s$ stand for meters and seconds respectively, and $m,n$ in the exponents stand for numbers from the initial formula 𝑣² = 𝑘𝑎^𝑚𝑆^𝑛.

Answer. $[k] = m^{2-n-m}s^{2m-2}$.