/*! 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 53 \(\bullet$$\bullet\) An object i... [FREE SOLUTION] | 91Ó°ÊÓ

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\(\bullet$$\bullet\) An object is dropped from rest and encounters air resist- ance that is proportional to the square of its speed. Sketch qualitative graphs (no numbers) showing (a) the air resistance on this object as a function of its speed, (b) the net force on the object as a function of its speed, (c) the net force on the object as a function of time, (d) the speed of the object as a function of time, and (e) the acceleration of the object as a function of time.

Short Answer

Expert verified
Graphs: (a) Parabola upward; (b & c) Downward curve to zero; (d) Asymptotic increase to constant; (e) Decrease to zero.

Step by step solution

01

Understanding Air Resistance

Air resistance is proportional to the square of speed, so if speed is denoted as \( v \), the air resistance \( R \) can be written as \( R = k v^2 \), where \( k \) is some constant of proportionality. This indicates a nonlinear increase in resistance with speed.
02

Graphing Air Resistance vs Speed

Since air resistance \( R = k v^2 \), the graph of air resistance as a function of speed will be a parabola opening upwards, starting from origin. As speed increases, air resistance increases sharply.
03

Analyzing Net Force vs Speed

Net force \( F_{net} \) is given by \( F_{net} = mg - kv^2 \) (where \( m \) is mass and \( g \) is gravitational acceleration). As speed increases, \( kv^2 \) increases, decreasing \( F_{net} \), resulting in a downward curve starting from \( mg \) on the force axis.
04

Graphing Net Force vs Speed

The graph will start at \( F = mg \) when \( v = 0 \) (object at rest) and will curve downward toward zero (terminal velocity), where gravity is balanced by air resistance.
05

Examining Net Force vs Time

Initially, \( F_{net} = mg \). As speed and air resistance increase over time, \( F_{net} \) decreases. Eventually reaching zero when terminal velocity is achieved.
06

Graphing Net Force vs Time

The graph starts high at \( F = mg \), then decreases asymptotically toward zero as time increases, reflecting the approach to terminal velocity.
07

Understanding Speed vs Time

Starting from rest, the speed increases as the net force decreases over time. Speed approaches a constant (terminal velocity) where net force becomes zero.
08

Graphing Speed vs Time

The speed graph starts from zero, increasing rapidly at first, then gradually less so, asymptotically approaching a constant terminal velocity.
09

Examining Acceleration vs Time

Acceleration \( a \) is defined as \( a = F_{net}/m \). Initially, \( a = g \), but decreases over time as the net force decreases, eventually reaching zero at terminal velocity.
10

Graphing Acceleration vs Time

The acceleration graph begins at \( g \) when the object is dropped and smoothly decreases to zero, reflecting the decrease in net force as the object reaches terminal velocity.

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Key Concepts

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

Understanding Net Force
Net force is the sum of all forces acting on an object. When an object falls, it encounters two major forces: gravity and air resistance.
Gravity pulls the object downward, while air resistance pushes against it.
The net force can be described by the equation \( F_{\text{net}} = mg - kv^2 \), where \( mg \) is the gravitational force and \( kv^2 \) is the air resistance force. Here, \( m \) represents mass, \( g \) is the acceleration due to gravity, \( v \) is velocity, and \( k \) is a constant.
  • When the object is just dropped, \( F_{\text{net}} = mg \) because speed \( v = 0 \), making air resistance zero.
  • As speed increases, air resistance \( kv^2 \) increases too, reducing \( F_{\text{net}} \).
  • Eventually, \( kv^2 \) equals \( mg \), making \( F_{\text{net}} = 0 \). This is when terminal velocity is reached.
Understanding this interaction helps explain how an object accelerates as it falls, while air resistance increasingly limits its speed.
Reaching Terminal Velocity
Terminal velocity is the constant speed that a falling object achieves once the net force on it is zero. This occurs when air resistance equals the force of gravity, creating a balanced force scenario.
The equation used is \( F_{\text{net}} = mg - kv^2 = 0 \), solving that gives \( mg = kv^2 \).
  • Terminal velocity depends on the balance between gravity \( mg \) and air resistance \( kv^2 \).
  • Since no net force acts on the object at terminal velocity, it stops accelerating and continues to move at a constant speed.
  • Different shapes and masses of objects reach terminal velocity at varying speeds due to differences in surface area and mass, which affect air resistance and gravitational pull.
By understanding how terminal velocity works, we see why skydivers reach a stable speed during free-fall without continuing to speed up indefinitely.
Deciphering Acceleration
Acceleration describes how quickly an object's speed changes. For an object in free-fall encountering air resistance, acceleration is directly linked to the net force acting on it.
The formula for acceleration is \( a = \frac{F_{\text{net}}}{m} \). This highlights how net force affects acceleration.
  • Initially, when the object is just dropped, acceleration \( a = g \) because net force \( F_{\text{net}} = mg \).
  • As the object speeds up, air resistance builds, reducing \( F_{\text{net}} \), and thus decreasing acceleration below \( g \).
  • At terminal velocity, net force becomes zero, causing acceleration to become zero too.
Understanding acceleration helps explain how objects transition from falling freely under the force of gravity, to moving at a constant speed when air resistance balances gravity.

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Most popular questions from this chapter

\(\bullet\) A light spring having a force constant of 125 \(\mathrm{N} / \mathrm{m}\) is used to pull a 9.50 \(\mathrm{kg}\) sled on a horizontal frictionless ice rink. If the sled has an acceleration of \(2.00 \mathrm{m} / \mathrm{s}^{2},\) by how much does the spring stretch if it pulls on the sled (a) horizontally, (b) at \(30.0^{\circ}\) above the horizontal?

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