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Answer :
The shortest time to pull a person out of a 38.8 m deep cave using a rope that can withstand 570 N is approximately 15.6 seconds.
To determine the shortest time in which a person can be pulled up from a 38.8 m deep cave using a rope, we need to analyze the forces and apply the kinematic equations. The person's weight is 522 N, and the rope can withstand a maximum tension of 570 N without breaking.
This is determined by calculating the maximum allowable acceleration and applying kinematic equations. The person's mass and the maximum tension the rope can withstand are essential factors in this calculation.
First, we'll find the person's mass using their weight:
m = F_g / g, where F_g = 522 N and g = 9.8 m/s²
m = 522 N / 9.8 m/s² = 53.27 kg
Next, we need to calculate the maximum allowable acceleration (a_max). The maximum tension (T_max) the rope can withstand is 570 N, and the actual tension (T) in the rope when lifting the person must be:
T = m(g + a)
For the maximum tension of 570 N:
570 N = 53.27 kg (9.8 m/s² + a_max)
Solving for a_max:
570 N = 553.05 N + 53.27 kg * a_max
570 N - 553.05 N = 53.27 kg * a_max
16.95 N = 53.27 kg * a_max
a_max = 16.95 N / 53.27 kg = 0.318 m/s²
Now, using the kinematic equation to find the shortest time (t) to cover the 38.8 m distance starting from rest:
d = 1/2 * a * t²
38.8 m = 1/2 * 0.318 m/s² * t²
38.8 m = 0.159 m/s² * t²
t² = 38.8 m / 0.159 m/s² = 244.03 s²
t = √(244.03 s²) ≈ 15.6 s
Therefore, the shortest time in which the person can be brought out of the cave is approximately 15.6 seconds.
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