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Question
Find the percentage error in kinetic energy of a body having mass 60.0 ± 0.3 g moving with a velocity of 25.0 ± 0.1 cm/s.
Solution
Given: m = 60.0 g, v = 25.0 cm/s, Δm = 0.3 g, Δv = 0.1 cm/s
To find: Percentage error in K.E.
Formula: K.E. = `1/2` mv2
Calculation: K.E. = `1/2 "mv"^2`
= `(Delta "K.E.")/("K.E.") xx 100`
`= ((triangle"m")/"m" + (2 * triangle"v")/"v") xx 100%`
= `(0.3/60.0 + (2 xx 0.1)/25.0) xx 100%`
= 1.3 %
The percentage error in kinetic energy is 1.3%.
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