If the error in measurements of mass and velocity is 1% each then error in measurement of kinetic energy is:

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Solution

Kinetic energy is $K = \frac{1}{2} m v^{2}$

Percentage error in kinetic energy

$= (\frac{Δ m}{m} + \frac{2 Δ v}{v}) \times 100$

$= (0.01 + 2 (0.01)) \times 100$

$= 0.03 \times 100$

$= 3$ %

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If the error in measurements of mass and velocity is 1% each then error in measurement of kinetic energy is:

Kinetic energy is K=12mv2 Percentage error in kinetic energy =(Δmm+2Δvv)×100 =(0.01+2(0.01))×100 =0.03×100 =3%

Kinetic energy is $K = \frac{1}{2} m v^{2}$

Percentage error in kinetic energy

$= (\frac{Δ m}{m} + \frac{2 Δ v}{v}) \times 100$

$= (0.01 + 2 (0.01)) \times 100$

$= 0.03 \times 100$

$= 3$ %