Margie Marx

2021-12-14

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?

### Answer & Explanation

Jonathan Burroughs

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
$\sum {F}_{y}=T-mg=m{a}_{y}=-3.0×6$
$T-mg=-18$
$T=18+mg=-18+\left(6.0×9.8\right)=40.8N$
$T=40.8N$

xleb123

Result:
$T=-40.8\phantom{\rule{0.167em}{0ex}}\text{N}$
Solution:
$F=m·a$
The net force acting on the bucket is the tension in the rope minus the weight of the bucket. When the bucket is being pulled up at a constant speed, the net force is zero, so the tension equals the weight:
$T-mg=0$
Substituting the given values:
$T-\left(6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\frac{m}{{s}^{2}}\right)=0$
$T-58.8\phantom{\rule{0.167em}{0ex}}\text{N}=0$
$T=58.8\phantom{\rule{0.167em}{0ex}}\text{N}$
Therefore, the tension in the rope when the bucket has an upward constant acceleration of magnitude $3\frac{m}{{s}^{2}}$ is approximately $78\phantom{\rule{0.167em}{0ex}}\text{N}$.
Now, let's find the tension in the rope when the bucket has a downward constant acceleration of magnitude $3\frac{m}{{s}^{2}}$. In this case, the net force acting on the bucket is the tension plus the weight of the bucket:
$T+mg=ma$
Substituting the given values:
$T+\left(6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\frac{m}{{s}^{2}}\right)=6\phantom{\rule{0.167em}{0ex}}\text{kg}×3\frac{m}{{s}^{2}}$
$T+58.8\phantom{\rule{0.167em}{0ex}}\text{N}=18\phantom{\rule{0.167em}{0ex}}\text{N}$
$T=18\phantom{\rule{0.167em}{0ex}}\text{N}-58.8\phantom{\rule{0.167em}{0ex}}\text{N}$
$T=-40.8\phantom{\rule{0.167em}{0ex}}\text{N}$
Therefore, the tension in the rope when the bucket has a downward constant acceleration of magnitude $3\frac{m}{{s}^{2}}$ is approximately $-40.8\phantom{\rule{0.167em}{0ex}}\text{N}$. Note that the negative sign indicates that the tension is directed downward.

Jazz Frenia

Let's start with the first question: At a certain point, the bucket starts accelerating upwards with a constant acceleration of magnitude $3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$. We need to find the tension in the rope in this scenario.
Initially, when the bucket was being pulled up at a constant speed, the tension in the rope was equal to the weight of the bucket. The weight is given by the formula $F=m×g$, where $m$ is the mass of the bucket and $g$ is the acceleration due to gravity ($9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$).
So, the tension on the string initially was:
$F=m×g=6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}=58.8\phantom{\rule{0.167em}{0ex}}\text{N}.$
Now, as the bucket starts accelerating upwards, there are two forces acting on it: the tension force in the rope and its weight. We can write the net force equation as:
$T-mg=ma,$
where $T$ is the tension in the rope and $a$ is the acceleration of the bucket.
Substituting the known values, we have:
$T-6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}=6\phantom{\rule{0.167em}{0ex}}\text{kg}×3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}.$
Simplifying this equation gives us:
$T=6\phantom{\rule{0.167em}{0ex}}\text{kg}×3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}+6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}=18\phantom{\rule{0.167em}{0ex}}\text{N}+58.8\phantom{\rule{0.167em}{0ex}}\text{N}=76.8\phantom{\rule{0.167em}{0ex}}\text{N}.$
Therefore, the tension in the rope when the bucket has an upward constant acceleration of magnitude $3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$ is approximately $76.8\phantom{\rule{0.167em}{0ex}}\text{N}$, which can be rounded to about $78\phantom{\rule{0.167em}{0ex}}\text{N}$.
Moving on to the second question: Now, let's consider the scenario where the bucket has a downward acceleration of magnitude $3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$. We need to find the tension in the rope in this situation.
The setup is similar to the previous question, but this time the direction of acceleration is reversed. The net force equation becomes:
$T+mg=ma.$
Substituting the known values, we have:
$T+6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}=6\phantom{\rule{0.167em}{0ex}}\text{kg}×\left(-3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right).$
Simplifying this equation gives us:
$T=6\phantom{\rule{0.167em}{0ex}}\text{kg}×\left(-3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}\right)-6\phantom{\rule{0.167em}{0ex}}\text{kg}×9.8\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}=-18\phantom{\rule{0.167em}{0ex}}\text{N}-58.8\phantom{\rule{0.167em}{0ex}}\text{N}=-76.8\phantom{\rule{0.167em}{0ex}}\text{N}.$
Therefore, the tension in the rope when the bucket has a downward constant acceleration of magnitude $3\phantom{\rule{0.167em}{0ex}}{\text{m/s}}^{2}$ is approximately $-76.8\phantom{\rule{0.167em}{0ex}}\text{N}$, which can be rounded to about $-78\phantom{\rule{0.167em}{0ex}}\text{N}$. The negative sign indicates that the tension is acting in the opposite direction (downward) as compared to the previous scenario.

fudzisako

Step 1. Upward acceleration:
When the bucket has an upward constant acceleration of 3 m/s², the net force acting on it is the sum of the gravitational force and the tension force. Let's denote the tension force as T.
Using Newton's second law, F = ma, we can write the equation of motion for the bucket as:
T - mg = ma
Here, m is the mass of the bucket (6 kg), g is the acceleration due to gravity (9.8 m/s²), and a is the upward acceleration (3 m/s²).
Substituting the values into the equation, we get:
T - (6 kg)(9.8 m/s²) = (6 kg)(3 m/s²)
Simplifying the equation:
T - 58.8 N = 18 N
To find the tension force T, we can rearrange the equation:
T = 18 N + 58.8 N = 76.8 N
Therefore, the tension in the rope when the bucket has an upward acceleration of 3 m/s² is approximately 76.8 N.
Step 2. Downward acceleration:
When the bucket has a downward constant acceleration of 3 m/s², the net force acting on it is again the sum of the gravitational force and the tension force.
Using Newton's second law as before, we have:
T - mg = ma
In this case, the acceleration a is negative (-3 m/s²) because it is directed downward.
Substituting the values into the equation:
T - (6 kg)(9.8 m/s²) = (6 kg)(-3 m/s²)
Simplifying the equation:
T - 58.8 N = -18 N
Rearranging the equation to find T:
T = -18 N + 58.8 N = 40.8 N
Therefore, the tension in the rope when the bucket has a downward acceleration of 3 m/s² is approximately 40.8 N.
In summary:
- When the bucket has an upward acceleration of 3 m/s², the tension in the rope is approximately $76.8N$.
- When the bucket has a downward acceleration of 3 m/s², the tension in the rope is approximately $40.8N$.

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