/*! 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 12 Alice drops a rock off a cliff. ... [FREE SOLUTION] | 91Ó°ÊÓ

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Alice drops a rock off a cliff. Bubba shoots a gun straight down from the edge of the same cliff. Compare the accelerations of the rock and the bullet while they are in the air on the way down. [Based on a problem by Serway and Faughn.]

Short Answer

Expert verified
Both the rock and the bullet have the same gravitational acceleration of about 9.81 m/s².

Step by step solution

01

Understanding the Problem

Alice drops a rock off a cliff, and Bubba shoots a bullet straight down from the same cliff. We need to compare the accelerations of both the rock and the bullet while they are in the air, moving towards the ground.
02

Identifying Forces Acting

Both the rock and the bullet, while in free fall, are subject only to the force of gravity, assuming air resistance is negligible. This implies that they are both only accelerated by gravitational force.
03

Applying Newton’s Second Law

Newton’s Second Law tells us that the force acting on both objects is the product of mass and acceleration. While the masses of the rock and the bullet may differ, gravitational acceleration is constant and is the same for any object under only the force of gravity.
04

Recognizing Gravitational Acceleration

The gravitational acceleration is approximately \(9.81 \text{ m/s}^2\) on Earth, which applies to all objects in free-fall near Earth's surface, regardless of their mass.
05

Conclusion about Accelerations

Since both the rock and the bullet are only under the influence of gravity, they both have the same gravitational acceleration of approximately \(9.81 \text{ m/s}^2\).

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Key Concepts

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

Newton's Second Law
Newton's Second Law is a fundamental principle that explains how the motion of an object is affected by forces. According to this law, the force applied to an object is equal to the mass of the object multiplied by its acceleration. This relationship is written in the equation form as \( F = ma \).

To understand this better, let's unpack it:
  • **Force (\( F \))**: This refers to any push or pull that can cause an object to accelerate.
  • **Mass (\( m \))**: The amount of matter in the object, which determines its resistance to acceleration when a force is applied.
  • **Acceleration (\( a \))**: The rate at which an object changes its velocity.
In the scenario with the rock and the bullet, Newton's Second Law helps us understand that even though their masses might differ, the gravitational force acting on them results in the same acceleration. This is because gravitational acceleration on Earth is a constant value, approximately \( 9.81 \, \text{m/s}^2 \). This means that, under the influence of only gravity, both objects experience the same acceleration downward regardless of their mass differences.
Free Fall
Free fall refers to the motion of an object under the influence of gravitational force only. When objects are in free fall, they are accelerating towards the Earth without the influence of other forces, such as air resistance.

Here are some important aspects of free fall:
  • **Gravitational Acceleration**: All objects in free fall on Earth experience the same gravitational acceleration of approximately \( 9.81 \, \text{m/s}^2 \).
  • **Negligible Air Resistance**: In idealized situations, we assume that air resistance is negligible, so the only force acting on the objects is gravity.
  • **Uniform Acceleration**: Since the gravitational force is constant, objects in free fall have a constant acceleration.
In the case of Alice's rock and Bubba's bullet, both experience free fall. Despite being a heavier or a lighter object, the gravitational pull accelerates them equally at \( 9.81 \, \text{m/s}^2 \), making their accelerations in free fall identical.
Force of Gravity
The force of gravity is the natural force that attracts two bodies toward each other, most notably observed between the Earth and objects near it. This force gives weight to physical objects and is a central concept in understanding motions involving gravitational acceleration.

Key points about the force of gravity include:
  • **Gravitational Force (\( F_g \))**: This is the force with which the Earth attracts an object towards its center and is calculated as \( F_g = mg \), where \( m \) is mass and \( g \) is gravitational acceleration.
  • **Equal Acceleration**: Regardless of the mass of the object, the gravitational force imparts an equal acceleration on all objects falling freely towards Earth.
  • **Weight**: The weight of an object is the result of the gravitational force acting on its mass. It is given by the same formula \( F_g = mg \).
When Alice drops her rock and Bubba shoots his bullet downward, both are subjected to the force of gravity. This is why they both have the same rate of acceleration during their descent, aligning with classical physics' understanding that gravitational pull affects all objects equally, leading to uniform acceleration in free fall.

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Most popular questions from this chapter

You're an astronaut, and you've arrived on planet \(X\), which is airless. You drop a hammer from a height of \(1.00 \mathrm{~m}\) and find that it takes \(350 \mathrm{~ms}\) to fall to the ground. What is the acceleration due to gravity on planet X?

The photo shows Apollo 16 astronaut John Young jumping on the moon and saluting at the top of his jump. The video footage of the jump shows him staying aloft for \(1.45\) seconds. Gravity on the moon is \(1 / 6\) as strong as on the earth. Compute the height of the jump. \(\sqrt{ }\)

A person is parachute jumping. During the time between when she leaps out of the plane and when she opens her chute, her altitude is given by an equation of the form $$y=b-c\left(t+k e^{-t / k}\right)$$ where \(e\) is the base of natural logarithms, and \(b, c\), and \(k\) are constants. Because of air resistance, her velocity does not increase at a steady rate as it would for an object falling in vacuum. (a) What units would \(b, c\), and \(k\) have to have for the equation to make sense? (b) Find the person's velocity, \(v\), as a function of time. [You will need to use the chain rule, and the fact that \(\mathrm{d}\left(e^{x}\right) / \mathrm{d} x=e^{x}\).] (c) Use your answer from part (b) to get an interpretation of the constant \(c\). [Hint: \(e^{-x}\) approaches zero for large values of \(x\).] (d) Find the person's acceleration, \(a\), as a function of time. (e) Use your answer from part (d) to show that if she waits long enough to open her chute, her acceleration will become very small.

In college-level women's softball in the U.S., typically a pitcher is expected to be at least \(1.75 \mathrm{~m}\) tall, but Virginia Tech pitcher Jasmin Harrell is \(1.62 \mathrm{~m}\). Although a pitcher actually throws by stepping forward and swinging her arm in a circle, let's make a simplified physical model to estimate how much of a disadvantage Harrell has had to overcome due to her height. We'll pretend that the pitcher gives the ball a constant acceleration in a straight line, and that the length of this line is proportional to the pitcher's height. Compare the acceleration Harrell would have to supply with the acceleration that would suffice for a pitcher of the nominal minimum height, if both were to throw a pitch at the same speed.

If an object starts accelerating from rest, we have \(v^{2}=\) \(2 a \Delta x\) for its speed after it has traveled a distance \(\Delta x .\) Explain in words why it makes sense that the equation has velocity squared, but distance only to the first power. Don't recapitulate the derivation in the book, or give a justification based on units. The point is to explain what this feature of the equation tells us about the way speed increases as more distance is covered.

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