The minimum time necessary to pull out Margo is 2.64 seconds.
Given the data in the question;
- Tension; [tex]T = 755N[/tex]
- mass; [tex]m = 70.0kg[/tex]
- depth or distance; [tex]s = 3.4m[/tex]
There are two forces acting on Margo, as shown in the image below, the rope Tension acting upward and the force of gravity acting downward.
From Newton's second law of motion:
[tex]F = m*a[/tex]
where F is the Force acting on the body, m is the mass and a is the acceleration.
Also, [tex]F = T - mg[/tex]
So, [tex]T - mg = ma[/tex]
We know that the value of "g" gravitation due to gravity is [tex]9.81m/s^2[/tex] and [tex]1Newton = 1 kg.m/s^2[/tex]
We substitute in our values to find "a"
[tex]755 kg.m/s^2 - ( 70.0kg*9.81m/s^2) = (70kg * a )[/tex]
[tex]755 kg.m/s^2 - 686.7 kg.m/s^2 = (70kg * a )[/tex]
[tex]68.3kg.m/s^2 = (70kg * a )[/tex]
[tex]a = \frac{68.3kg.m/s^2}{70kg}[/tex]
[tex]a = 0.9757 m/s^2[/tex]
Now, we set [tex]v_0 = 0[/tex] to get the minimum time required
From the second equation of motion:
[tex]s = v_0t + \frac{1}{2} at^2[/tex]
Since [tex]v_0 = 0[/tex]
[tex]s = \frac{1}{2} at^2[/tex]
We make time "t", the subject of the formula
[tex]t = \sqrt{\frac{2s}{a} }[/tex]
We substitute in our values
[tex]t = \sqrt{\frac{2*3.4m}{0.9757m/s^2} }[/tex]
[tex]t = 2.63995 s\\\\t = 2.64s[/tex]
Therefore, the minimum time necessary to pull out Margo is 2.64 seconds.
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