/*! 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 \(23 \mathrm{~... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 8: Problem 42

A worker pushed a \(23 \mathrm{~kg}\) block \(8.4 \mathrm{~m}\) 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) The work done by the worker is approximately 416.2 J. (b) The increase in thermal energy of the block-floor system is 416.2 J.

Step by step solution

01

Calculate the Normal Force

The block has a weight due to gravity given by \( W = mg \), where \( m = 23 \, \text{kg} \) and \( g = 9.8 \, \text{m/s}^2 \). This gives us \( W = 225.4 \, \text{N} \). However, because the worker applies a force at an angle, the vertical component affects the normal force \( N \). If \( F \) is the force applied by the worker, then the vertical component is \( F_y = F \sin(32^\circ) \). The normal force is \( N = mg - F_y \).
02

Calculate Frictional Force

The frictional force \( f_{k} \) is given by \( f_{k} = \mu_{k} N \), where \( \mu_{k} = 0.20 \). Substitute the normal force from Step 1 to find \( f_{k} = 0.20 \times (mg - F \sin(32^\circ)) \).
03

Determine Total Force by Worker

For constant speed, the horizontal component of the worker's force must equal the frictional force, thus \( F\cos(32^\circ) = f_{k} \). Using this equation, solve for \( F \) using values from Step 2.
04

Calculate Work Done by the Worker

Work done by the worker \( W_{worker} \) is calculated using \( W = Fd\cos(\theta) \), where \( d = 8.4\, \text{m} \) and \( \theta = 32^\circ \). Using the force \( F \) obtained in Step 3, calculate \( W_{worker} = F \times 8.4 \cos(32^\circ) \).
05

Calculate Increase in Thermal Energy

The increase in thermal energy \( \Delta E_{thermal} \) in the block-floor system is equal to the work done against friction, which is the frictional force multiplied by the distance: \( \Delta E_{thermal} = f_{k} \times 8.4 \).

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.

Work and Energy
Work and energy are two core concepts in physics. To start with, work is done whenever a force moves something over a distance. It鈥檚 calculated using the formula \( W = Fd \cos(\theta) \), where \( W \) is work, \( F \) is the force applied, \( d \) is the distance moved, and \( \theta \) is the angle between the force and the direction of movement. In this exercise, the worker pushes a block, applying force at an angle, which makes this formula handy.
Energy, closely connected to work, is the capacity to perform work. In this context, the energy used by the worker is converted into work done on the block and some amount heats up due to friction, which we see later.
Remember, whenever you do work on an object, not all energy goes into moving it forward. Some energy might transform into other forms, like heat, due to forces like friction.
Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding past each other. This is the resistance you overcome when sliding an object across a surface.
  • It's determined by the coefficient of kinetic friction, \( \mu_k \), which is unique for each surface combo.
  • The frictional force \( f_k \) is calculated as \( f_k = \mu_k N \), where \( N \) is the normal force.
In our problem, the normal force is not just the weight but is adjusted by the angle of the worker's push. The worker's force has a vertical component that changes the normal force.
This frictional force is significant because the worker must apply enough force to overcome it to keep the block moving at constant speed. It also plays a vital role in calculating the increase in the system's thermal energy.
Newton's Laws of Motion
To fully grasp this exercise, Newton's Laws of Motion are essential, particularly the first law, which states that an object remains in motion at a constant speed and direction if not acted upon by a net force. Here鈥檚 where understanding the motion of the block becomes intuitive.
In the given scenario:
  • The block moves at constant speed, implying that the net force is zero.
  • This means the force applied by the worker (in the horizontal direction) exactly matches the resisting frictional force.
  • The vertical component of the force also slightly reduces the normal force, affecting friction.
Newton鈥檚 second law, \( F = ma \), also subtly appears, highlighting that because the acceleration is zero (constant speed), the forces must balance out. Such a balance of forces allows us to establish the equations we use to figure out the specific values needed for solving the steps in the exercise.

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

At a lunar base, a uniform chain hangs over the edge of a horizontal platform. A machine does \(1.0 \mathrm{~J}\) of work in pulling the rest of the chain onto the platform. The chain has a mass of \(2.0\) \(\mathrm{kg}\) and a length of \(3.0 \mathrm{~m}\). What length was initially hanging over the edge? On the Moon, the gravitational acceleration is \(1 / 6\) of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\).

You push a \(2.0 \mathrm{~kg}\) block against a horizontal spring, compressing the spring by \(12 \mathrm{~cm}\). Then you release the block, and the spring sends it sliding across a tabletop. It stops \(75 \mathrm{~cm}\) from where you released it. The spring constant is \(170 \mathrm{~N} / \mathrm{m}\). What is the block-table coefficient of kinetic friction?

A single conservative force \(F(x)\) acts on a \(1.0 \mathrm{~kg}\) particle that moves along an \(x\) axis. The potential energy \(U(x)\) associated with \(F(x)\) is given by $$ U(x)=-4 x e^{-x / 4} \mathrm{~J} $$ where \(x\) is in meters. At \(x=5.0 \mathrm{~m}\) the particle has a kinetic energy of \(2.0 \mathrm{~J}\). (a) What is the mechanical energy of the system? (b) Make a plot of \(U(x)\) as a function of \(x\) for \(0 \leq x \leq 10 \mathrm{~m}\), and on the same graph draw the line that represents the mechanical energy of the system. Use part (b) to determine (c) the least value of \(x\) the particle can reach and (d) the greatest value of \(x\) the particle can reach. Use part (b) to determine (e) the maximum kinetic energy of the particle and (f) the value of \(x\) at which it occurs. (g) Determine an expression in newtons and meters for \(F(x)\) as a function of \(x\). (h) For what (finite) value of \(x\) does \(F(x)=0\) ?

A \(700 \mathrm{~g}\) block is released from rest at height \(h_{0}\) above a vertical spring with spring constant \(k=450 \mathrm{~N} / \mathrm{m}\) and negligible mass. The block sticks to the spring and momentarily stops after compressing the spring \(19.0 \mathrm{~cm}\). How much work is done (a) by the block on the spring and (b) by the spring on the block? (c) What is the value of \(h_{0} ?\) (d) If the block were released from height \(2.00 h_{0}\) above the spring, what would be the maximum compression of the spring?

Figure 8-23 shows a thin rod, of length \(L=2.00 \mathrm{~m}\) and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass \(m=5.00 \mathrm{~kg}\) is attached to the other end. The rod is pulled aside to angle \(\theta_{0}=30.0^{\circ}\) and released with initial velocity \(\vec{v}_{0}=0\). As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and (b) what is the change in the gravitational potential energy of the ball-Earth system? (c) If the gravitational potential energy is taken to be zero at the lowest point, what is its value just as the ball is released? (d) Do the magnitudes of the answers to (a) through (c) increase, decrease, or remain the same if angle \(\theta_{0}\) is increased?
See all solutions

Recommended explanations on Physics Textbooks

Atoms and Radioactivity

Read Explanation

Oscillations

Read Explanation

Space Physics

Read Explanation

Thermodynamics

Read Explanation

Rotational Dynamics

Read Explanation

Modern 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.