/*! 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 65 A pickup truck is carrying a t... [FREE SOLUTION] | 91Ó°ÊÓ

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A pickup truck is carrying a toolbox, but the rear gate of the truck is missing, so the box will slide out if it is set moving. The coefficients of kinetic and static friction between the box and the bed of the truck are 0.355 and \(0.650,\) respectively. Starting from rest, what is the shortest time in which this truck could accelerate uniformly to \(30.0 \mathrm{~m} / \mathrm{s}(\approx 60 \mathrm{mi} / \mathrm{h}) \quad\) without causing the box to slide? Include a free-body diagram of the toolbox as part of your solution. (Hint: First use Newton's second law to find the maximum acceleration that static friction can give the box, and then solve for the time required to reach \(30.0 \mathrm{~m} / \mathrm{s}\).)

Short Answer

Expert verified
The shortest time is approximately 4.70 seconds.

Step by step solution

01

Analyze Forces and Draw the Free-Body Diagram

Start by identifying the forces acting on the toolbox. The forces include the gravitational force downward, the normal force upward, and the static frictional force that opposes the sliding of the box. The static frictional force acts in the horizontal direction resisting movement. The free-body diagram will show these: Weight (mg), Normal force (N), and Static friction force (f_s).
02

Calculate Maximum Static Friction Force

The maximum static frictional force (f_s_max) can be calculated using the formula: \( f_{s_{ ext{max}}} = \mu_s imes N \). Since the box is on a level surface, the normal force (N) equals the gravitational force (mg). So, \( f_{s_{ ext{max}}} = 0.650 imes mg \).
03

Apply Newton's Second Law

According to Newton's second law, the maximum force due to static friction will provide the maximum acceleration the toolbox can have without sliding. This is stated as \( f_{s_{ ext{max}}} = ma \). Therefore, \( a = \frac{f_{s_{ ext{max}}}}{m} \). Substituting the expression for \( f_{s_{ ext{max}}} \), we have: \( a = 0.650g \).
04

Solve for Maximum Acceleration

Plugging in the acceleration due to gravity \( g = 9.81 \, \mathrm{m/s^2} \) gives: \( a = 0.650 \times 9.81 \approx 6.377 \, \mathrm{m/s^2} \). This is the maximum possible acceleration without sliding the box.
05

Calculate Shortest Time to Reach 30 m/s

Using the formula for constant acceleration \( v = at \), where \( v = 30.0 \, \mathrm{m/s} \), and \( a \) is the maximum acceleration, solve for \( t \). Hence, \( t = \frac{v}{a} = \frac{30.0}{6.377} \approx 4.70 \, \mathrm{seconds} \).

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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 fundamental to understanding how objects move. It tells us that the acceleration of an object is directly proportional to the net force acting on it. This can be highlighted with the equation:
  • \[ F = ma \]
Here \( F \) represents the net force applied, \( m \) is the mass of the object, and \( a \) is the acceleration produced.
Newton's Second Law is crucial in solving the toolbox problem. By knowing the maximum force that static friction can exert, we can determine how fast the truck can accelerate without the toolbox sliding. This boils down to finding that critical acceleration point where static friction is just enough to counteract any forces that might cause the toolbox to slide.
Understanding this principle allows us to calculate the maximum acceleration the box can withstand while remaining stationary relative to the truck's bed.
Static Friction
Static friction is the force that keeps an object at rest when it starts to experience an applied force. It acts parallel to the surfaces in contact:
  • Static friction holds steady until a force larger than its maximum can move the object.
  • The force depends on the contact surfaces and is usually described by the coefficient of static friction \( \mu_s \).
The formula for the static frictional force is:
  • \[ f_s = \mu_s \times N \]
where \( f_s \) is the static friction force, and \( N \) is the normal force, which balances the gravitational force in this scenario.
In the exercise, static friction prevents the toolbox from sliding off the truck bed. We need to calculate the maximum force that it can oppose. Once that force is known, the maximum acceleration without movement can be determined, effectively helping us solve the given problem.
Uniform Acceleration
Uniform acceleration means that the velocity of an object changes at a constant rate. This constant rate of acceleration simplifies calculating how time and distance relate:
  • Uniform acceleration includes scenarios where speed consistently increases or decreases.
  • It allows straightforward calculations using the formula:
  • \[ v = at \]
where \( v \) is the velocity, \( a \) is the acceleration, and \( t \) is the time taken.
In the toolbox problem, uniform acceleration is the method used by the truck to speed up to \( 30.0 \, \mathrm{m/s} \). Knowing the maximum acceleration helps us calculate the shortest time it will take to reach this speed using the above equation. This concept is central in balancing the forces with acceleration to ensure the toolbox doesn’t slide off.
Free-Body Diagram
A free-body diagram is a crucial tool in physics for visualizing how forces act on an object. It breaks down complex force problems:
  • Depicts all forces acting upon the object.
  • Indicates direction and type of each force (e.g., gravity, normal, frictional).
For our toolbox scenario:
  • Gravitational force (weight) pulls downward.
  • Normal force pushes upwards, balancing gravity.
  • Static friction force opposes potential horizontal movement.
Creating an accurate free-body diagram helps set up the problem for analysis using Newton's laws. It ensures a clear understanding of all acting forces, aiding in solving for unknowns like required acceleration or time. This diagram is instrumental in visualizing the forces at play and consequently solving for the time taken to reach a certain velocity without losing the box.

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Most popular questions from this chapter

Muscles are attached to bones by means of tendons. The maximum force that a muscle can exert is directly proportional to its cross-sectional area \(A\) at the widest point. We can express this relationship mathematically as \(F_{\max }=\sigma A\) where \(\sigma\) (sigma) is a proportionality constant. Surprisingly, \(\sigma\) is about the same for the muscles of all animals and has the numerical value of \(3.0 \times 10^{5}\) in SI units. The gastrocnemius muscle, in the back of the leg, has two portions, known as the medial and lateral heads. Assume that they attach to the Achilles tendon as shown in Figure \(5.45 .\) The cross sectional area of each of these two muscles is typically \(30 \mathrm{~cm}^{2}\) for many adults. What is the maximum tension they can produce in the Achilles tendon?

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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?

You've attached a bungee cord to a wagon and are using it to pull your little sister while you take her for a jaunt. The bungee's unstretched length is \(1.3 \mathrm{~m}\), and you happen to know that your little sister weighs \(220 \mathrm{~N}\) and the wagon weighs \(75 \mathrm{~N}\). Crossing a street, you accelerate from rest to your normal walking speed of \(1.5 \mathrm{~m} / \mathrm{s}\) in \(2.0 \mathrm{~s},\) and you notice that while you're accelerating, the bungee's length increases to about \(2.0 \mathrm{~m}\). What's the force constant of the bungee cord, assuming it obeys Hooke's law?

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