/*! 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 52 A 20.0 -kg rock is sliding on a ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 52

A 20.0 -kg rock is sliding on a rough, horizontal surface at 8.00 \(\mathrm{m} / \mathrm{s}\) and eventually stops due to friction. The coefficient of kinetic friction between the rock and the surface is \(0.200 .\) What average power is produced by friction as the rock stops?

Short Answer

Expert verified
The average power produced by friction is approximately 156.86 W.

Step by step solution

01

Identify Known Values

We are given the mass of the rock as \( m = 20.0 \) kg, the initial velocity \( v_i = 8.00 \) m/s, and the coefficient of kinetic friction \( \mu_k = 0.200 \). The rock eventually stops, so its final velocity \( v_f = 0 \) m/s.
02

Calculate Force of Friction

The force of kinetic friction \( f_k \) acting on the rock is calculated using the formula \( f_k = \mu_k \times m \times g \), where \( g = 9.81 \) m/s虏 is the acceleration due to gravity. Thus, \( f_k = 0.200 \times 20.0 \times 9.81 = 39.24 \) N.
03

Calculate Work Done by Friction

The work \( W \) done by friction as the rock comes to a stop is equal to the initial kinetic energy of the rock, which is \( W = \frac{1}{2} m v_i^2 = \frac{1}{2} \times 20.0 \times (8.00)^2 = 640.0 \) J.
04

Determine Time Duration for the Motion

The time \( t \) taken to stop the rock can be found using the work-energy principle and the definition of power: \( P_{avg} = \frac{W}{t} \). For uniform deceleration, the time \( t \) can be found through \( f_k = m \times a \), leading to \( a = \frac{f_k}{m} = \frac{39.24}{20.0} = 1.962 \) m/s虏. Using \( v_f = v_i - at \), solving for \( t \) gives \( t = \frac{v_i}{a} = \frac{8.00}{1.962} \approx 4.08 \) s.
05

Calculate Average Power

Finally, the average power \( P_{avg} \) produced by friction is calculated using \( P_{avg} = \frac{W}{t} = \frac{640.0}{4.08} \approx 156.86 \) W.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Kinetic Friction
When objects slide across each other, a special force resists their motion. This force is called kinetic friction. It acts opposite to the direction of movement and depends on both the surfaces in contact and their roughness.
  • **Coefficient of Kinetic Friction**: This is a unitless number, usually less than 1, representing how much two surfaces "grip" each other. For example, in our exercise, the coefficient of kinetic friction (\( \mu_k \)) is given as 0.200.
  • **Calculation of Kinetic Friction**: The frictional force (\( f_k \)) can be calculated using the formula: \( f_k = \mu_k \times m \times g \). Here, \( g \) is the acceleration due to gravity, approximately 9.81 m/s虏.
Understanding kinetic friction is crucial because it affects how quickly something can slow down or stop when sliding over a surface.
Work-Energy Principle
The work-energy principle is an important concept in physics. It states that the work done on an object is equal to the change in its kinetic energy. In simpler terms, the energy transferred through work can change an object鈥檚 speed or motion.
  • **Work Done by Forces**: When forces act on an object, they can do work. This work can increase or decrease the object鈥檚 energy. For instance, the frictional work (studied in our exercise) causes a moving object to slow down.
  • **Mathematical Expression**: Work (\( W \)) is expressed as the product of force, displacement, and the cosine of the angle between them. In our horizontal case of pure kinetic friction: \( W = f_k \times d \)
  • **Energy Changes**: The work causes a change in kinetic energy from its initial value to zero as the object stops.
The work-energy principle helps us understand how different forces can change an object's motion. It simplifies predicting how fast objects will slow down or stop.
Frictional Force
Frictional force is the resistance that one surface or object encounters when moving over another. It is a vital concept in physics and everyday life since it impacts motion.
  • **Types of Friction**: There are two main types: static friction (when objects are not moving) and kinetic friction (when objects are sliding past each other).
  • **Role in Motion**: In our exercise, the kinetic frictional force is the reason the rock comes to a stop.
    This force, acting in the opposite direction to the velocity, gradually reduces the rock's speed to zero.
  • **Calculation**: Kinetic frictional force (\( f_k \)) is determined using the coefficient of kinetic friction and the normal force (weight of the object in the absence of other vertical forces). In the example, it was calculated using: \( f_k = \mu_k \times m \times g \).
Frictional force is important for understanding how objects move and stop. It's a critical factor in vehicle braking, walks on icy roads, and many more real-life applications.
Kinetic Energy
Kinetic energy is the energy of motion. Any object that is moving has kinetic energy. This type of energy is directly linked to the speed of an object and its mass.
  • **Expression of Kinetic Energy**: The formula for kinetic energy (\( KE \)) is \( KE = \frac{1}{2} mv^2 \), where \( m \) represents mass and \( v \) is the velocity.
  • **Initial Kinetic Energy**: In our problem, the rock's initial kinetic energy was 640 J, calculated using its mass and initial velocity.
  • **Energy Conversion**: When the rock slides and gradually stops, its kinetic energy is transformed into other forms (like heat due to friction).
Kinetic energy is foundational in physics. It explains why objects move at certain speeds and how forces affect them during motion.
Average Power
Average power is a concept used to describe how quickly work is done over an interval of time. In physics, it helps us understand the rate of energy transfer.
  • **Definition**: Average power (\( P_{avg} \)) is the amount of work done or energy transferred per unit time, calculated as \( P_{avg} = \frac{W}{t} \).
  • **In the Exercise**: As shown in the example, the work done by the frictional force was used to calculate the average power as the rock came to a stop.
    Given the work and time (4.08 s) needed for the rock to stop, the average power produced by friction was approximately 156.86 W.
Average power explains how quickly energy was dissipated by friction. It's essential in engineering, mechanics, and many real-world applications that require energy calculations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Automotive Power I. A truck engine transmits 28.0 \(\mathrm{kW}(37.5 \mathrm{hp})\) to the driving wheels when the truck is traveling at a constant velocity of magnitude 60.0 \(\mathrm{km} / \mathrm{h}\) (37.3 \(\mathrm{mi} / \mathrm{h} )\) on a level road. (a) What is the resisting force acting on the truck? (b) Assume that 65\(\%\) of the resisting force is due to rolling friction and the remainder is due to air resistance. If the force of rolling friction is independent of speed, and the force of air resistance is proportional to the square of the speed, what power will drive the truck at 30.0 \(\mathrm{km} / \mathrm{h} ?\) At 120.0 \(\mathrm{km} / \mathrm{h} ?\) Give your answers in kilowatts and in horsepower.

A 2.0 -kg box and a 3.0 -kg box on a perfectly smooth horizontal floor have a spring of force constant 250 \(\mathrm{N} / \mathrm{m}\) compressed between them. If the initial compression of the spring is \(6.0 \mathrm{cm},\) find the acceleration of each box the instant after they are released. Be sure to include free-body diagrams of each box as part of your solution.

CALC A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let's try to get a reasonable approximation to this. Consider a spring of mass \(M,\) equilibrium length \(L_{0},\) and spring constant \(k .\) The work done to stretch or compress the spring by a distance \(L\) is \(\frac{1}{2} k X^{2}\) , where \(X=L-L-L_{0}\) . Consider a spring, as described above, that has one end fixed and the other end moving with speed \(v .\) Assume that the speed of points along the length of the spring varies linearly with distance \(l\) from the fixed end. Assume also that the mass \(M\) of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of the \(M\) and \(v .\) (Hint: Divide the spring into pieces of length \(d l ;\) find the speed of each pivide in terms of \(l, v,\) and \(L ;\) find the mass of each piece in terms of \(d l, M,\) and \(L ;\) and integrate from 0 to \(L .\) The result is \(n o t \frac{1}{2} M v^{2},\) since not all of the spring moves with the same speed.) In a spring gun, a spring of mass 0.243 \(\mathrm{kg}\) and force constant 3200 \(\mathrm{N} / \mathrm{m}\) is compressed 2.50 \(\mathrm{cm}\) from its unstretched length. When the trigger is pulled, the spring pushes horizon- tally on a 0.053 -kg ball. The work done by friction is negligible. Calculate the ball's speed when the spring reaches its uncom- pressed length (b) ignoring the mass of the spring and (c) includ- ing, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

CALC An airplane in flight is subject to an air resistance force proportional to the square of its speed \(v .\) But there is an additional resistive force because the airplane has wings. Air flowing over the wings is pushed down and slightly forward, so from Newton's third law the air exerts a force on the wings and airplane that is up and slightly backward (Fig. P6.104). The upward force is the lift force that keeps the airplane aloft, and the backward force is called induced drag. At flying speeds, induced drag is inversely proportional to \(v^{2},\) so that the total air resistance force can be expressed by \(F_{\text { air }}=\alpha v^{2}+\beta / v^{2},\) where \(\alpha\) and \(\beta\) are positive constants that depend on the shape and size of the airplane and the density of the air. For a Cessna \(150,\) a small single-engine airplane, \(\alpha=0.30 \mathrm{N} \cdot \mathrm{s}^{2} / \mathrm{m}^{2}\) and \(\beta=3.5 \times 10^{5} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{s}^{2} .\) In steady flight, the engine must provide a forward force that exactly balances the air resistance force. (a) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) at which this airplane will have the maximum range (that is, travel the greatest distance) for a given quantity of fuel. (b) Calculate the speed (in \(\mathrm{km} / \mathrm{h} )\) for which the airplane will have the maximum endurance (that is, remain in the air the longest time).

The spring of a spring gun has force constant \(k=400 \mathrm{N} / \mathrm{m}\) and negligible mass. The spring is compressed \(6.00 \mathrm{cm},\) and a ball with mass 0.0300 \(\mathrm{kg}\) is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 \(\mathrm{cm}\) long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so the barrel is horizontal. (a) Calculate the speed with which the ball leaves the barrel if you can ignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 \(\mathrm{N}\) acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel.)
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Modern Physics

Read Explanation

Oscillations

Read Explanation

Electric Charge Field and Potential

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Solid State Physics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.