The kinetic energy of a moving body is given by K = 1 2 mv^2 . If the mass of the body is measured as m = (2.0 0.1) kg and its velocity as v = (5.0 0.2) m/s, what is the maximum percentage error in the measurement of its kinetic energy?

Step-by-step solution

Given: m = (2.0 0.1) kg v = (5.0 0.2) m/s The percentage error in mass is: m m 100 = 0.1 2.0 100 = 5 % The percentage error in velocity is: v v 100 = 0.2 5.0 100 = 4 % The formula for kinetic energy is K = 1 2 mv^2 . The constant 1 2 does not contribute to the error. The maximum percentage error in K is: K K 100 = ( m m 100 ) + 2 ( v v 100 ) Substituting the values: K K 100 = 5 % + 2(4 %) = 5 % + 8 % = 13 % Answer: 13 %