Kinetic energy of a particle depends on the square of speed of the particle. If error in measurement of speed is $40\%$ then error in the measurement of kinetic energy will be

Hint : 

Error in kinetic energy is $\frac{\Delta K}{K}$.

$\text{Step1:Initial Kinetic Energy}$

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

$\mathrm{K}=1/2{\mathrm{m}\mathrm{v}}^2\dots .(1)$

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

$\text{Step2:Final Speed of particle}$

If error in speed is $40\%$

then new speed of particle $v^{\prime }=v+40\%\text{ of }v$

$=v+40v/100$

$=1.4\mathrm{v}$

$\text{Step3:Final Kinetic Energy}$

Final kinetic energy will be ${\mathrm{K}}^{\prime }=1/2{\mathrm{m}\mathrm{v}}^{\prime 2}$

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

$=1/2\mathrm{m}\left(1.96{\mathrm{v}}^2\right)$

$=(1.96)1/2m{v}^2$

$=\text{ (1.96) K }$  [ form equation (1), ]

$\text{Step4:Error in Kinetic Energy}$

Now, % error in Kinetic energy $=$ final kinetic energy$-$ initial kinetic energy$/$initial kinetic energy$\times 100$

$=\left({\mathrm{K}}^{\prime }-\mathrm{K}\right)/\mathrm{K}\times 100$

$=(1.96\mathrm{K}-\mathrm{K})/\mathrm{K}\times 100$

$=0.96\mathrm{K}/\mathrm{K}\times 100$

$=96\%$

Therefore maximum error in kinetic energy should be $96\%$.