Asked by ABC
The percentage errors in the measurement of mass and speed are 2% and 4%respectively. Then find maximum error in the measurement of K.E?
Answers
Answered by
Damon
KE = (1/2) m v^2
dKE= (v^2/2)dm + mvdv
dm = .02 m
dv = .04 v
dKE = (v^2/2) .02 m + m v .04 v
= m v^2/2 (.02) + mv^2/2 (.08 )
= 0.10 KE
= 10%
because velocity is squared, it alone counts for 2*4 = 8%
dKE= (v^2/2)dm + mvdv
dm = .02 m
dv = .04 v
dKE = (v^2/2) .02 m + m v .04 v
= m v^2/2 (.02) + mv^2/2 (.08 )
= 0.10 KE
= 10%
because velocity is squared, it alone counts for 2*4 = 8%