Question

calculate percentage error in kinetic energy of a body of mass 23 + - 0.1 g and moving with velocity of 46+_0.2 cm/s

Answer 1

Answer:

1.3 %.

Explanation:

Given:

• Mass of the body with uncertainty, $\rm m\pm \Delta m = 23\pm 0.1\ g.$

• Velocity of the body with uncertainty, $\rm v\pm \Delta v = 46\pm 0.2\ cm/s.$

Therefore,

$\rm m = 23\ g.\\\Delta m = 0.1\ g.\\v= 46\ cm/s.\\\Delta v = 0.2\ cm/s.$

The Kinetic energy of the body is given by

$\rm K.E. = \dfrac 12 mv^2.$

Therefore, the uncertainty in the kinetic energy of the body is given by

$\rm \dfrac{\Delta K.E.}{K.E.}=\dfrac{\Delta m}{m}+\dfrac{\Delta (v^2)}{v}\\\\\rm=\dfrac{\Delta m}{m}+\dfrac{2\Delta v}{v}\\\\=\dfrac{0.1}{23}+\dfrac{2\times 0.2}{46}\\=0.01304.\ \ \ ...........\ \ (1).$

The percentage error in the kinetic energy is given by

$\rm Percentage\ error =\dfrac{Change\ in\ K.E.}{Original\ K.E.}\times 100\%= \dfrac{\Delta K.E.}{K.E.}\times 100\%.$

Using (1),

$\rm Percentage\ error = 0.01304\times 100\% = 1.304\%\approx 1.3\%.$