Answer
$2.1×10⁻¹³ J$
Work Step by Step
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 $2.1\times10^{-13} J$.
