Answer
(a) The speed of the bucket is 1.80 m/s.
(b) The speed of the bucket must be at least 3.43 m/s.
Work Step by Step
(a) First we can find the net force acting on the bucket.
$\sum F = F_T - F_g$
$\sum F = 25.0 ~N - (2.00 ~kg)(9.80 ~m/s^2)$
$\sum F = 5.40 ~N$
We can use the net force to find the velocity.
$m\frac{v^2}{r} = \sum F$
$v^2 = \frac{(r)(\sum F)}{m}$
$v = \sqrt{\frac{(1.20 ~m)(5.40 ~N)}{(2.00 ~kg)}}$
$v = 1.80 ~m/s$
The speed of the bucket is 1.80 m/s.
(b) If the rope does not go slack, then the velocity at the top must be fast enough such that the centripetal force is equal to $F_g$ (which is $mg$).
$m\frac{v^2}{r} = mg$
$v^2 = gr$
$v = \sqrt{(9.80 ~m/s^2)(1.20 ~m)}$
$v = 3.43 ~m/s$
The speed of the bucket must be at least 3.43 m/s.
