/*! 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 42 A worker pushed a \(27 \mathrm{~... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 42

A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block- floor system?

Short Answer

Expert verified
(a) Calculate the work using \( W = Fd \cos \theta \). (b) Increase in thermal energy is \( \mu_k N \cdot d \).

Step by step solution

01

Identify Known Variables

First, identify and note down all the given variables:- Mass of the block, \( m = 27 \text{ kg} \).- Distance pushed, \( d = 9.2 \text{ m} \).- Angle of the applied force, \( \theta = 32^{\circ} \).- Coefficient of kinetic friction, \( \mu_k = 0.20 \).The gravitational acceleration \( g = 9.8 \text{ m/s}^2 \) is assumed as a known constant.
02

Calculate the Normal Force

Calculate the normal force. Since the force is below the horizontal, the vertical component of the force will affect the normal force.The normal force \( N \) can be calculated using:\[ N = mg - F \sin \theta \]Since the worker pushes the block at constant speed, there is no net horizontal force, so the horizontal component of the force balances with friction:\[ F \cos \theta = \mu_k N \]
03

Solve for the Applied Force

Use the horizontal force equation, \( F \cos \theta = \mu_k N \), and substitute the expression for normal force from the previous step:\[ F \cos \theta = \mu_k (mg - F \sin \theta) \]Solve for \( F \):\[ F \cos \theta + \mu_k F \sin \theta = \mu_k mg \]Rearrange to find \( F \):\[ F (\cos \theta + \mu_k \sin \theta) = \mu_k mg \]\[ F = \frac{\mu_k mg}{\cos \theta + \mu_k \sin \theta} \]
04

Calculate the Work Done by the Worker's Force

Once you have \( F \), calculate the work done by the worker's force:The formula for work done is:\[ W = Fd \cos \theta \]Substitute the known values and calculate \( W \).
05

Calculate the Increase in Thermal Energy

The increase in thermal energy of the block-floor system due to friction is equal to the work done against friction:\[ E_{\text{thermal}} = F_{\text{friction}} \cdot d \]Since \( F_{\text{friction}} = \mu_k N \), then:\[ E_{\text{thermal}} = \mu_k N \cdot d \]

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
Kinetic friction is a force that acts between moving surfaces. When a block slides across a floor, kinetic friction acts in the opposite direction of its motion. This force depends on:
  • The coefficient of kinetic friction (\( \mu_k \)), which is unique to the two surfaces in contact.
  • The normal force (\( N \)), which is the perpendicular force the surface exerts back on the block.
In the exercise, the kinetic friction force can be calculated by multiplying \( \mu_k \) by the normal force. This force is what causes the worker to apply effort to maintain constant speed. Without overcoming it, the block would slow down and stop due to resistance. Remember, kinetic friction does not depend on the speed of the sliding object, but it remains consistent across equal surfaces.
Work and Energy
Energy and work are crucial concepts in physics, especially when analyzing forces and movement. Work done by a force moving an object is calculated using:\[ W = Fd \cos \theta \]where \( W \) is the work done, \( F \) is the magnitude of the applied force, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of movement.
In this exercise, the work done includes the angle \( \theta \) since the force is applied below the horizontal. When work is done against friction, as in pushing a block, it often transforms into thermal energy, increasing the temperature at the interface.
It is vital to realize that this transformation is not a loss of energy but a conversion from mechanical work to thermal energy in the block-floor system. With constant forces and speeds, the work done equates to the energy dissipated as thermal energy.
Normal Force
The normal force is one of the essential forces acting against gravity. It is the force exerted by a surface to support the weight of an object resting on it and acts perpendicular to the surface.
The calculated normal force takes into account any additional vertical force components, such as when the applied force is not purely horizontal.
In the exercise, the normal force (\( N \)) impacts the amount of friction the block experiences. You can find it using the equation: \[ N = mg - F \sin \theta \]where \( mg \) is the gravitational force, and \( F \sin \theta \) is the vertical component of the applied force. Understanding how the normal force works helps in comprehending how much frictional force is at play, which is crucial for determining both the effort needed to move an object and the resulting energy changes.

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

From the edge of a cliff, a \(0.55 \mathrm{~kg}\) projectile is launched witl an initial kinetic energy of \(1550 \mathrm{~J}\). The projectile's maximum up ward displacement from the launch point is \(+140 \mathrm{~m}\). What are the (a) horizontal and (b) vertical components of its launch velocity? (c) At the instant the vertical component of its velocity is \(65 \mathrm{~m} / \mathrm{s}\), what is its vertical displacement from the launch point?

A \(70.0 \mathrm{~kg}\) man jumping from a window lands in an elevated fire rescue net \(11.0 \mathrm{~m}\) below the window. He momentarily stops when he has stretched the net by \(1.50 \mathrm{~m}\). Assuming that mechanical energy is conserved during this process and that the net functions like an ideal spring, find the elastic potential energy of the net when it is stretched by \(1.50 \mathrm{~m}\).

A horizontal force of magnitude \(35.0 \mathrm{~N}\) pushes a block of mass \(4.00 \mathrm{~kg}\) across a floor where the coefficient of kinetic friction is \(0.600\). (a) How much work is done by that applied force on the block-floon system when the block slides through a displacement of \(3.00 \mathrm{~m}\) across the floor? (b) During that displacement, the thermal energy of the block increases by \(40.0 \mathrm{~J}\). What is the increase in thermal energy of the floor? (c) What is the increase in the kinetic energy of the block?

A small block is sent through point \(A\) with a speed of \(7.0 \mathrm{~m} / \mathrm{s}\). Its path is without friction until it reaches the section of length \(L=12 \mathrm{~m}\), where the coefficient of kinetic friction is \(0.70 .\) The indicated heights are \(h_{1}=6.0 \mathrm{~m}\) and \(h_{2}=2.0 \mathrm{~m} .\) What are the speeds of the block at (a) point \(B\) and (b) point \(C ?\) (c) Does the block reach point \(D ?\) If so, what is its speed there; if not, how far through the section of friction does it travel?

A small block of mass \(m=0.032 \mathrm{~kg}\) can slide along the frictionless loop-the-loop, with loop radius \(R=12 \mathrm{~cm}\). The block is released from rest at point \(P\), at height \(h=5.0 R\) above the bottom of the loop. How much work does the gravitational force do on the block as the block travels from point \(P\) to (a) point \(Q\) and (b) the top of the loop? If the gravitational potential energy of the block-Earth system is taken to be zero at the bottom of the loop, what is that potential energy when the block is (c) at point \(P\), (d) at point \(Q\), and (e) at the top of the loop? (f) If, instead of merely being released, the block is given some initial speed downward along the track, do the answers to (a) through (e) increase, decrease, or remain the same?
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Quantum Physics

Read Explanation

Oscillations

Read Explanation

Fields in Physics

Read Explanation

Work Energy and Power

Read Explanation

Atoms and Radioactivity

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.