Answer:
The kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
Explanation:
Given;
initial velocity of proton, [tex]v_p_i[/tex] = 3 x 10⁵ m/s
distance moved by the proton, d = 3.5 m
electric field strength, E = 120 N/C
The kinetic energy of the proton at the end of the motion is calculated as follows.
Consider work-energy theorem;
W = ΔK.E
[tex]W =K.E_f - K.E_i[/tex]
where;
K.Ef is the final kinetic energy
W is work done in moving the proton = F x d = (EQ) x d = EQd
[tex]K.E_f =EQd + \frac{1}{2}m_pv_p_i^2[/tex]
[tex]m_p \ is \ mass \ of \ proton = 1.673 \ \times \ 10^{-27} kg \\\\Q \ is \ charge \ of \ proton = 1.6 \times 10^{-19} C[/tex]
[tex]K.E_f = 120\times 1.6 \times 10^{-19} \times 3.5 \ + \ \frac{1}{2}(1.673\times 10^{-27})(3\times 10^5)^2 \\\\[/tex]
[tex]K.E_f = 6.72\times 10^{-17} \ + \ 7.53 \times 10^{-17} \\\\K.E_f = 14.25 \times 10^{-17} J\\\\K.E_f = 1.425\times 10^{-16} \ J[/tex]
Therefore, the kinetic energy of the proton at the end of the motion is 1.425 x 10⁻¹⁶ J.
