The force on the small sphere due to the electric field created by the line of charge is given by:
$F = \frac{kq\lambda}{r}$
where k is Coulomb's constant, q is the charge on the small sphere, λ is the linear charge density of the line of charge, and r is the distance between the sphere and the line of charge.
At the initial position, r = 1.60 cm = 0.0160 m, and at the final position, r = 4.20 cm = 0.0420 m.
The work done by this electric force on the sphere as it moves from the initial position to the final position is:
$W = \int_{r_i}^{r_f} F dr = \int_{0.0160m}^{0.0420m} \frac{kq\lambda}{r} dr = kq\lambda\ln\frac{r_f}{r_i}$
where ln is the natural logarithm.
The work done by the electric force is equal to the change in kinetic energy of the sphere, so we have:
$\frac{1}{2}mv_f^2 - \frac{1}{2}mv_i^2 = kq\lambda\ln\frac{r_f}{r_i}$
Since the sphere starts from rest, vi = 0. Solving for vf, we get:
$v_f = \sqrt{\frac{2kq\lambda}{m}\ln\frac{r_f}{r_i}}$
where m is the mass of the sphere. To find the kinetic energy, we use:
$K = \frac{1}{2}mv_f^2$
The mass of the sphere is not given, so we cannot calculate the kinetic energy.