Question

percentage errors of measurement in
velocity and mass are 2% and 4% respectively, what is the percentage error in kinetic energy?

Answer 1

Method 1:
The kinetic energy of an object is given by the relation
T=12mv2

Now, lets forget about the units for a moment and take logarithm on both sides, we get :

lnT=−ln2+lnm+2lnv

Taking differential on both sides, we have the expression,

ΔTT=Δmm+2Δvv

This is the expression for fractional error and now if you multiply both sides with 100 you get percentage errors. So the answer to your question is

3 + 2*4 =11

Method 2:
From the expression for kinetic energy, take a small variation of it which can be expressed as

ΔT=12Δmv2+12m(2vΔv)

Now, divide both sides of this expression with our original expression for T, and again we get the old formula:

ΔTT=Δmm+2Δvv

Hope this is better. Its simple differential calculus.

Hope this helps you.
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