Newton's Second Law Equation:

$\overline{){\mathbf{\Sigma F}}{\mathbf{=}}{\mathbf{ma}}}$

**(a) **Forces acting on the bucket:

$\begin{array}{rcl}\mathbf{T}\mathbf{-}\mathbf{mg}& \mathbf{=}& \mathbf{ma}\\ \mathbf{T}& \mathbf{=}& \mathbf{m}\mathbf{g}\mathbf{+}\mathbf{m}\mathbf{a}\\ & \mathbf{=}& \mathbf{m}\mathbf{(}\mathbf{g}\mathbf{+}\mathbf{a}\mathbf{)}\end{array}$

- m = 6kg
- g = 9.8 m/s
^{2} - a = 3 m/s
^{2}

6 kg bucket of water is being pulled straight up by a string at a constant speed. I determined that the tension on the string was

F = ma

F = (6kg * 9.8 m/s^{2})

F = 58.8 N.

(a) At a certain point, the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude 3 m/s^{2}. What is the tension in the rope now?

(b) Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude 3 m/s^{2}. What is the tension in the rope?