Question

The percentage error in the mass and speed are 2% and 3% respectively. What is the maximum error in kinetic energy ? Calculate using these quantities?







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Answer 1

For a quantity Z defined as Z=XnYm where X,Y are physical quantities and n.m are real numbers, the percentage error in Z, that is,

$( \frac{Δz}{z})</p><p> \: is \: given \: by \\ ( \frac{Δz}{z} \%) = ( |n| \times \frac{Δx}{x} \%) + ( |m| \times \frac{Δy}{y} \%) \\ Where \: ( \frac{Δx}{x} \%) \: and \: ( \frac{Δy}{y} \%)$

are the percentage errors in X,Y respectively.

Kinetic energy KE of a body of mass m moving at a speed v is given by$KE = \frac{1}{2} {mv}^{2}$

we will first find the relation of kinetic energy with the mass and speed of the body and then use this relation to find a relation for the percentage error in the kinetic energy in terms of the percentage errors in the mass and speed of the body.

Hence, let us proceed to do that.

Kinetic energy KE of a body of mass m moving at a speed v is given by

$KE = \frac{1}{2} {mv}^{2}$

Also,

for a quantity Z defined as Z=XnYm where X,Y are physical quantities and n.m are real numbers, the percentage error in Z, that is,

$( \frac{Δz}{z} \%)$

is given by,

$( \frac{Δz}{z} \%) = ( |n| \times \frac{Δx}{x} \%) + ( |m| \times \frac{Δy}{y} \%)$

Hence, we get

$\frac{Δke}{ke} \: (in \: \%) = (1 \times \frac{Δ</p><p> \frac{1}{2} }{ \frac{1}{2} } \%) \\ + (1 \times \frac{Δ</p><p>m}{m} \%) \\ + (2 \times \frac{Δ</p><p>v}{v} \%)$

Now, according to the question, the percentage error in the mass and speed of the body are 2% and 3% respectively.

$∴( \frac{Δm}{m} \%) = 2\% \\ ∴( \frac{Δv}{v} \%) = 3\%$

Now put the 3rd equation together. Hence, the required percentage error in the kinetic energy of the body is 8%. Therefore the correct answer is 8%