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Error in kinetic energy is $KΔK $.

Kinetic energy of a particle depends on square of speed of a particle.

$K=1/2mv_{2}….(1)$

Let initial velocity of a particle is v and initial kinetic energy is $K$

$Step2:Final Speed of particle$

If error in speed is $40%$

then new speed of particle $v_{′}=v+40%ofv$

$=v+40v/100$

$=1.4v$

$Step3:Final Kinetic Energy$

Final kinetic energy will be $K_{′}=1/2mv_{′2}$

$=1/2m(1.4v)_{2}$

$=1/2m(1.96v_{2})$

$=(1.96)1/2mv_{2}$

$=(1.96) K$ [ form equation (1), ]

Now, % error in Kinetic energy $=$ final kinetic energy$−$ initial kinetic energy$/$initial kinetic energy$×100$

$=(K_{′}−K)/K×100$

$=(1.96K−K)/K×100$

$=0.96K/K×100$

$=96%$

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