Question

Kinetic energy of a particle depends on the square of speed of the particle if error in measurement of speed is 30% then the error in measurement of kinetic energy will be??
See the answer is 69 % and not 60 %. Please solve and show how come the answer is 69%.

Answer 1

Answer with Explanation:

let the measured KE and actual KE be $K_{m}$ and $K_{a}$

let the actual velocity and measured velocity be  v and  1.3 v [ as per given]

we know that

K $\alpha$ v²

$\frac{K_{m}}{K_{a}}=\frac{(1.3v)^2}{v^{2}}$

$K_{m} = 1.69 K_{a}$ ----------------------- ( 1 )

now,

percentage error = $\frac{observed value - correct value}{correct value} * 100 %$

percentage error in KE = $\frac{K_{m} - K_{a}}{K_{a}}*100$

                                     = $\frac{1.69 K_{a} -K_{a}}{K_{a}} * 100$

                                     = $\frac{(1.69 - 1) K_{a}}{K_a}*100$

                                    = 69 %

NOTE:

the method you used to calculate is applicable only when the percentage error is small ( > 5 %  or > 10%)

use this method for large errors

Hope it helps !

Answer 2

Answer with Explanation:

let the measured KE and actual KE be $K_{m}$ and $K_{a}$

let the actual velocity and measured velocity be v and 1.3 v [ as per given]

we know that

K $\alpha$ v²

$\frac{K_{m}}{K_{a}}$= $\frac{(1.3v)^2}{v^{2}}$

K_{m} = 1.69 K_{a} ----------------------- ( 1 )

now,

percentage error = $\frac{observed value - correct value}{correct value} * 100 %$

percentage error in KE = $\frac{K_{m} - K_{a}}{K_{a}}*100$

= $\frac{1.69 K_{a} -K_{a}}{K_{a}} * 100$

= $\frac{(1.69 - 1) K_{a}}{K_a}*100$

= 69 %

NOTE:

the method you used to calculate is applicable only when the percentage error is small ( > 5 % or > 10%)

use this method for large errors

Hope it helps !