Question

If the error in measurement of mass of a body be 3% and in the measurement of velocity be 2%. What will be maximum possible error in calculation of kinetic energy? (Kinetic Energy is given as K.E.= 1/2 mv2)

Answer 1

The kinetic energy of an object is given by the relation

T=12mv2

Take logarithm on both sides, we get :

lnT=−ln2+lnm+2lnv

Taking differential on both sides, we have the expression,

ΔTT=Δmm+2Δvv

Multiply both sides with 100 you get percentage errors, So:

3 + 2*4 =11

Answer 2

Answer:

The maximum possible error in kinetic energy is 7%.

Explanation:

Given that,

Percentage error in mass=3%

Percentage error in velocity=2%

The kinetic energy is defined as:

$K.E= \dfrac{1}{2}mv^2$

The percentage error of kinetic energy is

Total percentage error in kinetic energy = percentage error in mass+percentage error in velocity

Total percentage error in kinetic energy $K.E=3+2\times2$

Total percentage error in kinetic energy =7 %

Hence, The maximum possible error in kinetic energy is 7%.