 # A 6kg bucket of water is being pulled straight up by a strin Margie Marx 2021-12-14 Answered
A 6kg bucket of water is being pulled straight up by a string at a constant speed. The tension on the string was F = ma
$F=\left(6kg×9.8\frac{m}{s}2\right)$ , $F=58.8N$ Now its asking At a certain point the speed of the bucket begins to change. The bucket now has an upward constant acceleration of magnitude $3\frac{m}{{s}^{2}}$. What is the tension in the rope now? The correct answer was "about 78N" then Now assume that the bucket has a downward acceleration, with a constant acceleration of magnitude $3\frac{m}{{s}^{2}}$.What is the tension in the rope?
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Firstly, $\sum {F}_{y}=T-mg=m{a}_{y}=0$
As ${a}_{y}=0$, so $⇒T-mg=0$
$T=mg=6.0×9.8=58.8$
Now, use the same formula to find the tension in the rope.
Do the acceleration, when the bucket has an upward acceleration, it means that it has a positive value ${a}_{y}=+3\frac{m}{{s}^{2}}$
When it has a downward acceleration, it means that it has a negative value of ${a}_{y}=-3\frac{m}{{s}^{2}}$
Now,
$\sum {F}_{y}=T-mg=m{a}_{y}=+3.0×6$
$T-mg=18$
$T=18+mg=18+\left(6.0×9.8\right)=76.8N$
Now, you should apply the same thing into the downward acceleration
$\sum {F}_{y}=T-mg=m{a}_{y}=-3.0×6$
$T-mg=-18$
$T=18+mg=-18+\left(6.0×9.8\right)=40.8N$
$T=40.8N$