Answer:
(A) As it moves farther and farther from Q, its speed will keep increasing.
Explanation:
When a positive charge Q is fixed on a horizontal frictionless tabletop and a second charge q is released near to it then according to the Coulombs law the force acting on it decreases with the square of the distance between them.
Mathematically:

where:
r = distance between the charges
permittivity of free space
By the Newtons' second law of motion if the we know that the acceleration is directly proportional to the force applied. So as the distance between the charges increases the its acceleration also decreases therefore now the charge feels less acceleration but still continues to accelerate with a fading magnitude.
So your finding acceleration first which is 30m/s divides by 6 seconds equals 5m/s^s and then multiply that by 1,400 kg and you have net force which is 7,000N
From the geometry of the problem, the 20 m-long cable creates
the hypotenuse of a right triangle, with the extended of the other two sides of
size 20 m * cos(30 deg), which is around 17.3 m. Therefore, the ball has increased
by 20 m - 17.3 m = 2.7 m.
The potential energy will have altered by m*g*h, which is 1400 kg * 9.8 m/s^2 *
1.6 m , or about 37044 joules.