/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 The Fall of \(T\). rex Paleontol... [FREE SOLUTION] | 91Ó°ÊÓ

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The Fall of \(T\). rex Paleontologists estimate that if a Tynamosaurus rex were to trip and fall, it would have experienced a force of approximately \(260,000 \mathrm{N}\) acting on its torso when it hit the ground. Assuming the torso has a mass of \(3800 \mathrm{kg},\) (a) find the magnitude of the torso's upward acceleration as it comes to rest. (For comparison, humans lose consciousness with an acceleration of about \(7 g .\) (b) Assuming the torso is in free fall for a distance of \(1.46 \mathrm{m}\) as it falls to the ground, how much time is required for the torso to come to rest once it contacts the ground?

Short Answer

Expert verified
(a) The acceleration is approximately 68.42 m/s². (b) It takes about 0.206 seconds to come to rest.

Step by step solution

01

Calculate Acceleration

To find the acceleration, we use Newton's second law, which states: \[ F = m \cdot a \]where \( F \) is the force, \( m \) is the mass, and \( a \) is the acceleration. Rearrange the formula to solve for \( a \):\[ a = \frac{F}{m} \]Substitute the given values:\[ a = \frac{260,000 \text{ N}}{3800 \text{ kg}} \approx 68.42 \text{ m/s}^2 \]
02

Observing Comparison with Human Tolerance

The calculated acceleration is \( 68.42 \text{ m/s}^2 \). For comparison, humans experience loss of consciousness at around \( 7g \), where \( g = 9.8 \text{ m/s}^2 \). Therefore, for humans:\[ 7g = 7 \times 9.8 \approx 68.6 \text{ m/s}^2 \]\[ 68.42 \text{ m/s}^2 \] is slightly below \( 68.6 \text{ m/s}^2 \), near the threshold of human tolerance.
03

Find Time to Come to Rest

To find the time required for the torso to come to rest, use the kinematic equation:\[ v^2 = u^2 + 2a s \]Where \( v \) is the final velocity (0 m/s after stopping), \( u \) is the initial velocity, \( a \) is the acceleration, and \( s \) is the distance (1.46 m). Solving for initial velocity \( u \):\[ 0 = u^2 + 2(-68.42) \times 1.46 \]\[ u^2 = 2 \times 68.42 \times 1.46 \approx 199.438 \]\[ u \approx \sqrt{199.438} \approx 14.12 \text{ m/s} \]Now, use \( v = u + at \) to find time \( t \):\[ 0 = 14.12 + (-68.42) \cdot t \]\[ t = \frac{14.12}{68.42} \approx 0.206 \text{ seconds} \]

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Key Concepts

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

Force and Acceleration
Newton's Second Law provides a straightforward method for understanding how forces interact with mass to produce acceleration. Given by the formula \( F = m \cdot a \), this law allows us to solve for any unknown among the force \( F \), mass \( m \), or acceleration \( a \) when the other two are known. In our example of the toppled Tyrannosaurus rex, we use the given force of \( 260,000 \, \text{N} \) and the mass of \( 3800 \, \text{kg} \) to calculate a substantial upward acceleration of approximately \( 68.42 \, \text{m/s}^2 \). This calculation illustrates how significant forces can influence even massive objects to accelerate rapidly.

It's crucial to note how acceleration affects living organisms differently. While this magnitude of acceleration approaches the threshold of \( 68.6 \, \text{m/s}^2 \) where humans may lose consciousness, for a dinosaur, this would have marked a critical point in its fall. Understanding the interaction between forces and acceleration provides profound insights into motion dynamics, not only in physics but also in biological systems.
Kinematics
Kinematics, the study of motion without considering forces, plays an essential role in determining how objects move through space over time. Using kinematic equations, we can analyze the trajectory and time span of an object's movement. In our scenario with the Tyrannosaurus rex, once it started its descent and contacted the ground, we calculated how quickly it came to rest using the equation \( v^2 = u^2 + 2as \), where \( v \) is the final velocity, \( u \) is the initial velocity, \( a \) is acceleration, and \( s \) is the distance of 1.46 m.

From our calculations, the initial velocity right before the beast halted was found to be about \( 14.12 \, ext{m/s} \). With this known initial velocity and the upward acceleration calculated previously, we determined that it took approximately 0.206 seconds for the dinosaur's torso to come to a full stop. Through these calculations, kinematics enables us to precisely understand the timeline and final stopping distance of the motion.
Free Fall
The concept of free fall is central to understanding how gravity influences motion. In free fall, gravity is the only force acting on an object, causing it to accelerate downwards at a constant rate \( g \) of \( 9.8 \, \text{m/s}^2 \). In this exercise, we considered the T. rex's torso as being in free fall during its 1.46 m descent before impacting the ground, though the impact itself introduces other forces not covered under free fall principles.

When objects fall under the sole influence of gravity, movement can be straightforwardly predicted. However, when an object like the T. rex's torso hits the ground, external forces come into play, halting the fall and creating the need to apply other calculations. Understanding free fall is vital, though, as it sets the basis to comprehend how gravity alone acts on bodies before other forces modify their motion. By comparing these concepts, we appreciate how an object in free fall transitions to experiencing complex forces upon striking another surface.

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Most popular questions from this chapter

Give the direction of the net force acting on each of the following objects. If the net force is zero, state "zero." (a) A car accelerating northward from a stoplight. (b) A car traveling southward and slowing down. (c) A car traveling westward with constant speed. (d) A skydiver parachuting downward with constant speed. (e) A baseball during its flight from pitcher to catcher (ignoring air resistance).

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