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Question Number 126603 by Ar Brandon last updated on 22/Dec/20

If there is a positive error in the measurement of velocity of a body, then the error in the measure- ment of kinetic energy is

$$\mathrm{If}\:\mathrm{there}\:\mathrm{is}\:\mathrm{a}\:\mathrm{positive}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measurement} \\ $$$$\mathrm{of}\:\mathrm{velocity}\:\mathrm{of}\:\mathrm{a}\:\mathrm{body},\:\mathrm{then}\:\mathrm{the}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measure}- \\ $$$$\mathrm{ment}\:\mathrm{of}\:\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{is} \\ $$

Answered by Olaf last updated on 22/Dec/20

E = (1/2)mv^2 lnE = ln(1/2)+lnm+2lnv (dE/E) = 2(dv/v) ((ΔE)/E) = 2((Δv)/v) The relative error in the measurement of the kinetic energy is the double of the relative error in the measurement of the velocity

$$\mathrm{E}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{mv}^{\mathrm{2}} \\ $$$$\mathrm{lnE}\:=\:\mathrm{ln}\frac{\mathrm{1}}{\mathrm{2}}+\mathrm{ln}{m}+\mathrm{2ln}{v} \\ $$$$\frac{{d}\mathrm{E}}{\mathrm{E}}\:=\:\mathrm{2}\frac{{dv}}{{v}} \\ $$$$\frac{\Delta\mathrm{E}}{\mathrm{E}}\:=\:\mathrm{2}\frac{\Delta{v}}{{v}} \\ $$$$\mathrm{The}\:\mathrm{relative}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measurement} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{kinetic}\:\mathrm{energy}\:\mathrm{is}\:\mathrm{the}\:\mathrm{double}\:\mathrm{of} \\ $$$$\mathrm{the}\:\mathrm{relative}\:\mathrm{error}\:\mathrm{in}\:\mathrm{the}\:\mathrm{measurement} \\ $$$$\mathrm{of}\:\mathrm{the}\:\mathrm{velocity} \\ $$

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