Discussion
For a given mass m, linear velocity v, linear displacement s, linear acceleration a and time t, the dimensionally INCORRECT expression for power P is
Answer: C
The dimensionally incorrect expression is \(\frac{1}{2}mv^{2}\).
Power has dimensions \([ML^{2}T^{-3}]\). \(\frac{1}{2}mv^{2}\) has \([M][LT^{-1}]^{2} = [ML^{2}T^{-2}]\), the dimensions of kinetic energy, not power.
The others all reduce to \([ML^{2}T^{-3}]\):
- mav: \([M][LT^{-2}][LT^{-1}]\) (force times velocity)
- \(\frac{mv^{2}}{t}\): energy per unit time
- \(ma^{2}t\): \([M][L^{2}T^{-4}][T]\)
- \(\frac{mas}{t}\): work (force times displacement) per unit time
Numerical factors like \(\frac{1}{2}\) have no dimensions, so they do not decide the answer.
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