Answer:
The kinetic energy is [tex]K_f = 1.424 *10^{-15} \ J[/tex]
Explanation:
From the question we are told that
The initial speed of the proton is [tex]v_p = 3.0 * 10^{5} \ m/s[/tex]
The distance covered is [tex]d = 3.5 \ m[/tex]
The magnitude of the electric field is [tex]E = 120\ N/C[/tex]
Generally the mass of a proton is [tex]m = 1.67 *10^{27} \ kg[/tex]
and the charge on a proton is [tex]q= 1.60*10^{-19} \ C[/tex]
Now according to work energy theorem,
[tex]Work \ Done(W) = Kinetic \ Energy \ Change[/tex]
=> [tex]W = K_f -K_i[/tex]
=> [tex]K_f = W +K_i[/tex]
Where [tex]K_f[/tex] is final kinetic energy and [tex]K_i[/tex] is initial kinetic energy which is mathematically represented as
[tex]K_i = \frac{1}{2} * m * v_p^2[/tex]
Now the net workdone(W) is mathematically represented as
[tex]W = F * d = q* E* d[/tex]
So
[tex]K_f = q* E * d + \frac{1}{2} * m * v_p^2[/tex]
substituting values
[tex]K_f = 1.60*10^{-19}* 120 * 3.5 + \frac{1}{2} * 1.67*10^{-27} *(3.0*10^{5})^2[/tex]
[tex]K_f = 1.424 *10^{-15} \ J[/tex]
