/*! 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 20 A 5.0 -kg rock and a \(3.0 \time... [FREE SOLUTION] | 91Ó°ÊÓ

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A 5.0 -kg rock and a \(3.0 \times 10^{-4}\) -kg pebble are held near the surface of the earth. (a) Determine the magnitude of the gravitational force exerted on each by the earth. (b) Calculate the magnitude of the acceleration of each object when released.

Short Answer

Expert verified
(a) Rock: 49 N, Pebble: 2.94 x 10^{-3} N; (b) Both accelerate at 9.8 m/s².

Step by step solution

01

Define the Gravitational Force Formula

The gravitational force exerted on an object by the Earth is calculated using the formula: \[ F = m imes g \]where:- \( F \) is the gravitational force,- \( m \) is the mass of the object,- \( g \) is the acceleration due to gravity (approximately \( 9.8 \text{ m/s}^2 \) on the surface of the Earth).
02

Calculate the Gravitational Force on the Rock

Given:- Mass of the rock, \( m = 5.0 \text{ kg} \)- \( g = 9.8 \text{ m/s}^2 \)Use the formula:\[ F_{ ext{rock}} = 5.0 imes 9.8 \]Calculate:\[ F_{ ext{rock}} = 49 ext{ N} \]Thus, the gravitational force on the rock is \( 49 \text{ N} \).
03

Calculate the Gravitational Force on the Pebble

Given:- Mass of the pebble, \( m = 3.0 imes 10^{-4} \text{ kg} \)- \( g = 9.8 \text{ m/s}^2 \)Use the formula:\[ F_{ ext{pebble}} = 3.0 imes 10^{-4} imes 9.8 \]Calculate:\[ F_{ ext{pebble}} = 2.94 imes 10^{-3} ext{ N} \]Thus, the gravitational force on the pebble is \( 2.94 imes 10^{-3} \text{ N} \).
04

Define the Formula for Acceleration

The acceleration of an object when released is simply the acceleration due to gravity, \( g \), because no other forces act on it. Hence, for any object near Earth's surface, the acceleration is \( 9.8 \text{ m/s}^2 \).
05

Calculate the Acceleration for Both Objects

Since both objects are released near the Earth's surface, both the rock and the pebble will accelerate at the same rate when released:- Acceleration of the rock, \( a_{ ext{rock}} = g = 9.8 \text{ m/s}^2 \)- Acceleration of the pebble, \( a_{ ext{pebble}} = g = 9.8 \text{ m/s}^2 \)Both objects have the same acceleration due to gravity.

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Key Concepts

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

Mass and Weight
Mass and weight are often confused as the same concept, but they are distinct physical properties. Mass is the amount of matter in an object and is typically measured in kilograms (kg). It is an intrinsic property, meaning it does not change regardless of location. On the other hand, weight is the gravitational force exerted on an object's mass. This force depends on the object's mass and the gravitational acceleration where it is located. Therefore, weight varies with changes in gravity, while mass remains constant.

In the context of the exercise, the mass of the rock is 5.0 kg and of the pebble is \(3.0 \times 10^{-4}\) kg. The weight of each can be calculated using the formula, \( W = m \times g \), where \( W \) is the weight, \( m \) is the mass, and \( g \) is the acceleration due to gravity (approximately \( 9.8 \, \text{m/s}^2 \) on Earth).
The computed weights are 49 N for the rock and \(2.94 \times 10^{-3}\) N for the pebble. This illustrates how heavier objects (greater mass) exert a stronger gravitational force, or weight, while lighter objects exert less.
Acceleration Due to Gravity
Acceleration due to gravity, denoted as \( g \), is a crucial concept in physics. It represents the rate at which an object accelerates when falling freely under the influence of gravity alone. On Earth's surface, this acceleration is approximately \( 9.8 \, \text{m/s}^2 \). This constant value means that, barring air resistance, any object will accelerate at this rate when dropped.

In the given exercise, both the rock and the pebble, when released, accelerate at \( 9.8 \, \text{m/s}^2 \) towards the Earth. This acceleration is consistent regardless of their different masses. Intuitively, this might seem counterintuitive, as heavier objects feel like they should fall faster under gravity. However, in a vacuum, all objects indeed fall at the same rate, which is beautifully demonstrated by dropping a hammer and a feather on the Moon, where there is no air resistance.
Newton's Laws of Motion
Understanding Newton's laws of motion helps us grasp the fundamentals of forces and movement. The first law states that an object will remain at rest or continue in uniform motion unless acted upon by a net external force. This is known as inertia.

The second law links force, mass, and acceleration. It can be expressed as \( F = m \times a \), where \( F \) is the net force applied, \( m \) is mass, and \( a \) is acceleration. In the context of gravitational force, this transforms to \( F = m \times g \), illustrating how gravity impacts objects.

The third law states that for every action, there is an equal and opposite reaction. This law helps explain why objects exert reciprocal forces on each other, such as the rock and the Earth pulling on each other with equal force.
In the exercise, these laws explain why both the rock and the pebble fall toward Earth. While their masses differ, the gravitational force proportionally adjusts to keep their acceleration constant at \( 9.8 \, \text{m/s}^2 \). Newton's insights give us a robust framework to understand the motion of objects under various forces, including gravity.

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Most popular questions from this chapter

In the amusement park ride known as Magic Mountain Superman, powerful magnets accelerate a car and its riders from rest to 45 \(\mathrm{m} / \mathrm{s}\) (about 100 \(\mathrm{mi} / \mathrm{h} )\) in a time of 7.0 \(\mathrm{s}\) . The combined mass of the car and riders is \(5.5 \times 10^{3} \mathrm{kg}\) . Find the average net force exerted on the car and riders by the magnets.

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Scientists are experimenting with a kind of gun that may eventually be used to fire payloads directly into orbit. In one test, this gun accelerates a \(5.0-\mathrm{kg}\) projectile from rest to a speed of \(4.0 \times 10^{3} \mathrm{m} / \mathrm{s}\) . The net force accelerating the projectile is \(4.9 \times 10^{5} \mathrm{N}\) . How much time is required for the projectile to come up to speed?

ssm mm A rocket of mass \(4.50 \times 10^{5} \mathrm{kg}\) is in flight. Its thrust is directed at an angle of \(55.0^{\circ}\) above the horizontal and has a magnitude of \(7.50 \times 10^{6} \mathrm{N}\) . Find the magnitude and direction of the rocket's acceleration. Give the direction as an angle above the horizontal.

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