## Fundamentals of Physics Extended (10th Edition)

$2.1×10⁻¹³ J$
Let initial velocity of proton be $u=2.4\times10^{7} m/s$ (which is given) and velocity after travelling 3.5cm be $v$. So, increase in Kinetic Energy is $1/2m(v^{2}-u^{2})$. By equation of motion $v^{2}=u^{2}+2as$, we get the value of $v$: $v^{2}=(2.4\times10^{7})^{2}+2(3.6\times10^{15})0.035$ $v=2.9\times10^{7}$ So, change in Kinetic Energy is: $=1/2\times1.67\times10^{-27}((2.9× 10⁷)² −(2.4×10⁷)²)$ =$1/2\times1.67\times2.65\times10^{-13}$ =$2.212\times10^{-13}$ $\approx2.1\times10^{-13}$ Change in kinetic energy is $2.1\times10^{-13} J$.