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Chapter 4: Problem 75

A person whose weight is \(5.20 \times 10^{2} \mathrm{~N}\) is being pulled up vertically by a rope fron the bottom of a cave that is \(35.1 \mathrm{~m}\) deep. The maximum tension that the rope can withstand without breaking is \(569 \mathrm{~N}\). What is the shortest time, starting from rest, in which the person can be brought out of the cave?

Short Answer

Expert verified
The shortest time is approximately 8.72 seconds.

Step by step solution

01

Identify Forces

The forces acting on the person include the gravitational force (weight) downward, which is given as \(5.20 \times 10^2\, \mathrm{N}\), and the tension in the rope upward, which must be \(569 \mathrm{~N}\) without breaking. The net force \(F_{net}\) acting on the person is calculated as:\[ F_{net} = T - W = 569 - 520 = 49\, \mathrm{N} \]
02

Calculate Acceleration

According to Newton's second law, \(F = ma\), where \(F\) is the net force, \(m\) is the mass, and \(a\) is the acceleration. First, find the mass using the weight: \[ W = mg \Rightarrow m = \frac{W}{g} = \frac{520}{9.8} \approx 53.06 \mathrm{~kg}\]Then, calculate the acceleration:\[ a = \frac{F_{net}}{m} = \frac{49}{53.06} \approx 0.923 \mathrm{~m/s^2} \]
03

Apply Kinematic Equation

To find the shortest time, use the kinematic equation for motion with constant acceleration, starting from rest:\[ s = ut + \frac{1}{2} a t^2 \]Given that the initial velocity \(u = 0\), distance \(s = 35.1\, \mathrm{m}\), and \(a = 0.923 \mathrm{~m/s^2}\), this simplifies to:\[ 35.1 = \frac{1}{2} (0.923) t^2 \]Solving for \(t^2\):\[ t^2 = \frac{2 \times 35.1}{0.923} \approx 76.08 \]\[ t \approx \sqrt{76.08} \approx 8.72 \mathrm{~s} \]
04

Check Constraints

Check to ensure that this tension (569 N) does not exceed the maximum limit, which it does not. Thus, the constraints are satisfied with the calculated time.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

gravitational force
Gravitational force is a natural phenomenon by which objects with mass are attracted towards each other. On Earth, this force is typically experienced as weight. In the given exercise, the weight of the person being pulled out of the cave is provided as \(5.20 \times 10^{2} \mathrm{~N}\). This weight is essentially the gravitational force acting downward on the person.
The gravitational force is calculated using the equation \( W = mg \), where \( W \) is the weight, \( m \) is the mass of the object, and \( g \) is the acceleration due to Earth's gravity, approximately \(9.8 \mathrm{~m/s^2}\).
  • This force is always directed towards the center of the Earth.
  • It's essential for determining the overall force balance in any scenario involving vertical movement.
Understanding the gravitational pull is crucial when you need to calculate other forces and motions involved in lifting or holding an object against gravity.
tension
Tension refers to the pulling force exerted by a string, rope, cable, or similar object when it is stretched by forces acting at each end. In the scenario described, the rope's tension is what counteracts the gravitational force to lift the person.
The problem states that the maximum tension the rope can withstand is \(569 \mathrm{~N}\). If this amount is exceeded, the rope might break, hence ensuring that tension remains below this limit is crucial for safe retrieval. Tension acts in the opposite direction to the gravitational force, effectively keeping the person suspended or moving upwards.
  • Tension must always be considered in scenarios involving ropes or cables.
  • It must be balanced accurately against other forces to ensure effective and safe operation.
In our case, the tension provides the necessary force to overcome the gravitational force, assisting in calculating net force and subsequent acceleration.
kinematic equations
Kinematic equations are utilized to describe the motion of objects in terms of their position, velocity, and acceleration over time. These equations are particularly useful in cases involving constant acceleration, like the person being pulled from the bottom of a cave.
In the exercise, we use the kinematic equation \( s = ut + \frac{1}{2} a t^2 \) to determine the shortest time needed for the task. Here:
  • \( s \) is the distance traveled, \( 35.1 \mathrm{~m} \).
  • \( u \) is the initial velocity, which is \(0\) since the person starts from rest.
  • \( a \) is the acceleration we found to be \(0.923 \mathrm{~m/s^2}\).
  • \( t \) is the time, the unknown variable we solve for.
The kinematic equations allow us to determine different components of motion and are invaluable tools in physics for predicting and analyzing trajectories, especially when objects start from rest or are subject to constant forces.

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Most popular questions from this chapter

Concept Questions Two skaters, a man and a woman, are standing on ice. Neglect any friction between the skate blades and the ice. The woman pushes on the man with a certain force that is parallel to the ground. (a) Must the man accelerate under the action of this force? If so, what three factors determine the magnitude and direction of his acceleration? (b) Is there a corresponding force exerted on the woman? If so, where does it originate? Is this force related to the magnitude and direction of the force the woman exerts on the man? If so, how? Problem The mass of the man is \(82 \mathrm{~kg}\) and that of the woman is \(48 \mathrm{~kg} .\) The woman pushes on the man with a force of \(45 \mathrm{~N}\) due east. Determine the acceleration (magnitude and direction) of (a) the man and (b) the woman.

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