**Hint : **

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

**$$\textbf{Step1:Initial Kinetic Energy}$$**

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

$$\mathrm{K}=1 / 2 \mathrm{mv}^{2} \ldots .(1)$$

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

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

If error in speed is $$40 \%$$

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

$$=v+40 v / 100$$

$$=1.4 \mathrm{v}$$

$$\textbf{Step3:Final Kinetic Energy}$$

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

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

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

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

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

**$$\textbf{Step4:Error in Kinetic Energy}$$**

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 ****$$96 \%$$.**