To determine the SI unit of power, let’s consider what power represents in physics. Power is defined as the rate at which work is done or energy is transferred over time.
The formula for power [tex]\( P \)[/tex] is:
[tex]\[ P = \frac{W}{t} \][/tex]
where [tex]\( W \)[/tex] is work and [tex]\( t \)[/tex] is time.
Work ([tex]\( W \)[/tex]), in turn, is defined as:
[tex]\[ W = F \cdot d \][/tex]
where [tex]\( F \)[/tex] is force and [tex]\( d \)[/tex] is displacement.
We know that force ([tex]\( F \)[/tex]) itself can be expressed as:
[tex]\[ F = m \cdot a \][/tex]
where [tex]\( m \)[/tex] is mass and [tex]\( a \)[/tex] is acceleration.
Acceleration ([tex]\( a \)[/tex]) is defined as:
[tex]\[ a = \frac{\Delta v}{\Delta t} \][/tex]
which, considering it in terms of distance (displacement) and time, simplifies to:
[tex]\[ a = \frac{d}{t^2} \][/tex]
By substituting the expression for acceleration back into the force equation, we get:
[tex]\[ F = m \cdot \frac{d}{t^2} \][/tex]
Now, substituting this into the work equation:
[tex]\[ W = m \cdot \frac{d}{t^2} \cdot d \][/tex]
[tex]\[ W = m \cdot \frac{d^2}{t^2} \][/tex]
Finally, putting this value of work into the power formula:
[tex]\[ P = \frac{m \cdot \frac{d^2}{t^2}}{t} \][/tex]
[tex]\[ P = m \cdot \frac{d^2}{t^3} \][/tex]
From the above, we see that the SI unit of power is derived from the units of mass ([tex]\( kg \)[/tex]), distance ([tex]\( m \)[/tex]), and time ([tex]\( s \)[/tex]) as follows:
[tex]\[ P = kg \cdot m^2 \cdot s^{-3} \][/tex]
Therefore, the correct SI unit of power is:
[tex]\[ kg \cdot m^2 \cdot s^{-3} \][/tex]
This is not listed among the given choices:
a) [tex]$kg \cdot m \cdot s^{-2}$[/tex]
b) [tex]$kg \cdot m^2 \cdot s^{-2}$[/tex]
Thus, the correct SI unit for power is [tex]\( kg \cdot m^2 \cdot s^{-3} \)[/tex], which is not an option provided in the question.
