/*! 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 82 Consult Interactive LearningWare... [FREE SOLUTION] | 91Ó°ÊÓ

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Consult Interactive LearningWare 4.2 at before beginning this problem. A truck is traveling at a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) along a level road. A crate is resting on the bed of the truck, and the coefficient of static friction between the crate and the truck bed is \(0.650 .\) Determine the shortest distance in which the truck can come to a halt without causing the crate to slip forward relative to the truck.

Short Answer

Expert verified
The shortest stopping distance is approximately 49.00 meters.

Step by step solution

01

Understand the problem

We need to find the shortest distance in which a truck can stop without a crate sliding forward. The truck's initial speed is given, and we'll use the static friction to determine the deceleration.
02

Identify known values

Initial speed of the truck, \(v_0 = 25.0 \text{ m/s}\), coefficient of static friction, \(\mu_s = 0.650\), acceleration due to gravity, \(g = 9.81 \text{ m/s}^2\).
03

Calculate maximum static friction force

The static frictional force can be found using \(f_s = \mu_s \cdot m \cdot g\). However, since frictional force will be equal to the mass times acceleration, \(f_s = m \cdot a\), mass \(m\) cancels out when solving for acceleration (\(a\)).
04

Set up the equation for maximum deceleration

We equate the static friction force to mass times acceleration: \(f_s = m \cdot a = \mu_s \cdot m \cdot g\). Simplifying, we get \(a = \mu_s \cdot g\).
05

Substitute known values to find deceleration

\(a = 0.650 \cdot 9.81 \text{ m/s}^2 = 6.3765 \text{ m/s}^2\). This is the maximum deceleration that prevents the crate from sliding.
06

Use kinematic formula to find stopping distance

The formula is \(v^2 = v_0^2 + 2a \cdot d\). Here \(v = 0\) (final speed), \(a = -6.3765 \text{ m/s}^2\), \(v_0 = 25.0 \text{ m/s}\). Rearrange to find \(d\): \(d = \frac{v_0^2}{2|a|}\).
07

Calculate stopping distance

Substitute \(v_0 = 25.0\), and \(|a| = 6.3765\) into the equation: \(d = \frac{(25.0)^2}{2 \times 6.3765} = \frac{625}{12.753} \approx 49.00 \text{ m}\).

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Key Concepts

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

Static Friction
Static friction plays a crucial role in preventing objects from slipping or sliding. It refers to the force exerted by a surface to keep an object at rest. In the context of our exercise, the truck is attempting to stop, and the static friction between the crate and the truck bed is what prevents the crate from sliding forward as the truck decelerates.
The static friction force can be calculated using the formula:
  • \( f_s = \mu_s \cdot m \cdot g \)
Here:
  • \( f_s \) is the static friction force
  • \( \mu_s \) is the coefficient of static friction (a measure of how easily the surfaces can slide over each other)
  • \( m \) is the mass of the object
  • \( g \) is the acceleration due to gravity
In our problem, the mass of the crate cancels out when we use it to calculate the deceleration needed to keep the crate in place. This means that static friction helps dictate how much the truck can safely slow down without causing any movement in the crate.
Kinematics Equations
Kinematics equations are used to describe the motion of objects. They allow us to calculate distances, velocities, and acceleration over time. For our exercise, we are particularly interested in the equation that relates velocity, acceleration, and distance:
  • \( v^2 = v_0^2 + 2a \cdot d \)
In this formula:
  • \( v \) is the final velocity
  • \( v_0 \) is the initial velocity
  • \( a \) is the acceleration
  • \( d \) is the distance
As the truck's final velocity will be zero (since it comes to a stop), the equation simplifies to finding how far the truck can travel (\(d\)) before stopping. This is valuable because it gives us a way to calculate stopping distances required to prevent the crate from moving.
Deceleration Calculation
Deceleration is a term used to describe negative acceleration, where an object slows down. The objective in our exercise was to find the maximum deceleration that the static friction can support.
  • The maximum deceleration was determined using the relationship \( a = \mu_s \cdot g \).
Here, \( a \) stands for deceleration, and it equates to the product of the static friction coefficient and the gravitational acceleration.
This approach shows the importance of the static friction coefficient in determining how quickly an object can be slowed down before it starts moving.
By plugging in the known values of static friction and gravity, we found that the maximum deceleration was \( 6.3765 \text{ m/s}^2 \). This result was used in our kinematic equation to find how the truck, and consequently the crate, can safely be brought to a stop without any slipping.

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Most popular questions from this chapter

Three forces act on a moving object. One force has a magnitude of \(80.0 \mathrm{~N}\) and is directed due north. Another has a magnitude of \(60.0 \mathrm{~N}\) and is directed due west. What must be the magnitude and direction of the third force, such that the object continues to move with a constant velocity?

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Refer to Concept Simulation \(4.4\) at for background relating to this problem. The drawing shows a large cube (mass \(=25 \mathrm{~kg}\) ) being accelerated across a horizontal frictionless surface by a horizontal force \(\vec{P}\). A small cube \((\mathrm{mass}=4.0 \mathrm{~kg})\) is in contact with the front surface of the large cube and will slide downward unless \(\vec{P}\) is sufficiently large. The coefficient of static friction between the cubes is \(0.71\). What is the smallest magnitude that \(\overrightarrow{\mathrm{P}}\) can have in order to keep the small cube from sliding downward?

A \(1380-\mathrm{kg}\) car is moving due east with an initial speed of \(27.0 \mathrm{~m} / \mathrm{s}\). After \(8.00 \mathrm{~s}\) the car has slowed down to \(17.0 \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the net force that produces the deceleration.

A duck has a mass of \(2.5 \mathrm{~kg}\). As the duck paddles, a force of \(0.10 \mathrm{~N}\) acts on it in a direction due east. In addition, the current of the water exerts a force of \(0.20 \mathrm{~N}\) in a direction of \(52^{\circ}\) south of east. When these forces begin to act, the velocity of the duck is \(0.11 \mathrm{~m} / \mathrm{s}\) in a direction due east. Find the magnitude and direction (relative to due east) of the displacement that the duck undergoes in \(3.0 \mathrm{~s}\) while the forces are acting.

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