Question

the percentage error in the mass and speed are 2%and3% respectively. what is maximum error is kinetic energy​

Answer 1

Explanation:

Kinetic Energy,=1/2 mv^2

∆m=2%

∆v=3%

(∆KE/KE)=(∆m/m) x 2(∆v/v)

∆KE/KE= (2) +2(3)=8%

Answer 2

Answer:

$\huge { \underline\red {\bf{Solution}}}$

Explanation:

Given percentage errors in mass and velocity = 2% and 3% respectively

The kinetic energy of the particle is given by

$KE = 2mv2</p><p>$

The percentage error in kinetic energy will be

$mΔm×100+2vΔv×100</p><p></p><p></p><p>$

$= 2+6$

$= 8%$