Question

if there is a positive error of 50% in the measurement of velocity of a body then the error in the measurement of kinetic energy is

Answer 1

let the velocity with out error be v then velocity with error be 1.5v. So error% in kinetic energy=1/2m((1.5v)^2)-1/2 mv^2 /(1/2mv^2) ×100
2.25v^2-v^2 /v^2 =1.25×100=125%.

Answer 2

Let the velocity without error be v then velocity with error be 1.5v.

So error% in kinetic energy=1/2m((1.5v)^2)-1/2 mv^2 /(1/2mv^2) ×100 2.25v^2-v^2 /v^2 =1.25×100=125%.

This should be identified with error and positive errors are taking place in right measurement of the body.

It highlights in kinetic energy and total value can be measured.