/*! 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 7 A factory worker moves a \(30.0 ... [FREE SOLUTION] | 91Ó°ÊÓ

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A factory worker moves a \(30.0 \mathrm{~kg}\) crate a distance of \(4.5 \mathrm{~m}\) along a level floor at constant velocity by pushing horizontally on it. The coefficient of kinetic friction between the crate and the floor is 0.25 . (a) What magnitude of force must the worker apply? (b) How much work is done on the crate by the worker's push? (c) How much work is done on the crate by friction? (d) How much work is done by the normal force? By gravity? (e) What is the net work done on the crate?

Short Answer

Expert verified
(a) 73.5 N applied force. (b) 330.75 J by worker. (c) -330.75 J by friction. (d) 0 J by normal and gravity. (e) Net work is 0 J.

Step by step solution

01

Analyzing the Forces Involved

First, identify the forces acting on the crate: the gravitational force (weight), the normal force, the force of kinetic friction, and the applied force by the worker. Since the crate is moving at constant velocity, the net force is zero.
02

Calculating the Normal Force

Since there is no vertical movement, the normal force equals the gravitational force. The gravitational force can be calculated as:\[ F_{gravity} = m \cdot g = 30.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 294 \, \text{N} \] Hence, the normal force is also 294 N.
03

Finding the Force of Kinetic Friction

The force of kinetic friction is given by:\[ F_{friction} = \mu_k \cdot F_{normal} = 0.25 \times 294 \, \text{N} = 73.5 \, \text{N} \]
04

Calculating the Applied Force by the Worker

Since the crate is moving at constant velocity, the applied force equals the frictional force:\[ F_{applied} = F_{friction} = 73.5 \, \text{N} \]
05

Calculating Work Done by the Worker

Work done by the applied force is:\[ W_{worker} = F_{applied} \times d = 73.5 \, \text{N} \times 4.5 \, \text{m} = 330.75 \, \text{J} \]
06

Calculating Work Done by Friction

Work done by friction is also given by:\[ W_{friction} = -F_{friction} \times d = -73.5 \, \text{N} \times 4.5 \, \text{m} = -330.75 \, \text{J} \]The work is negative because the force of friction opposes the motion.
07

Work Done by the Normal Force and Gravity

The normal force and gravitational force act perpendicularly to the direction of the motion. Thus, the work done by both forces is zero:\[ W_{normal} = 0 \, \text{J} \]\[ W_{gravity} = 0 \, \text{J} \]
08

Calculating the Net Work Done on the Crate

The net work done is the sum of all works:\[ W_{net} = W_{worker} + W_{friction} + W_{normal} + W_{gravity} = 330.75 \, \text{J} - 330.75 \, \text{J} + 0 \, \text{J} + 0 \, \text{J} = 0 \, \text{J} \] The net work done is zero, consistent with the fact that there is no change in kinetic energy (constant velocity).

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Key Concepts

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

Kinetic Friction
Kinetic friction is a crucial concept in physics, especially when you need to understand how objects move across surfaces. In this exercise, kinetic friction plays a pivotal role as it opposes the motion of the crate being pushed by the factory worker. Friction is the force that resists motion, and when an object is moving, we refer to this resistance as kinetic friction.

The strength of kinetic friction depends on two factors: the nature of the surfaces in contact and how much they are pressed together. This is often quantified using the coefficient of kinetic friction (\( \mu_k \)), a dimensionless number that characterizes the sliding nature between two surfaces. In our scenario, the force of kinetic friction is calculated using the formula:\[F_{friction} = \mu_k \cdot F_{normal}\]Where \( F_{normal} \) is the normal force that acts perpendicular to the surface.
  • Kinetic friction arises between the crate and the floor.
  • Its magnitude must be overcome by any applied force to keep the crate moving.
Understanding kinetic friction helps us recognize why more force might be needed to move heavier objects.
Constant Velocity
Constant velocity implies that an object is moving at a steady speed and in a straight line. A critical aspect of this condition is that the net force acting on the object must be zero.

In this exercise, the crate is moved at constant velocity, which means that the force applied by the worker matches the force of kinetic friction exactly. No acceleration occurs because all forces are balanced.
  • Forces that oppose movement, like friction, must be balanced by equal forces that promote movement.
  • If the velocity changes, the balance between forces would shift, indicating non-zero net work.
Achieving constant velocity is why the worker's applied force equals the friction force. Recognizing constant velocity situations is key to understanding equilibrium in physics.
Force Calculation
Calculating forces correctly is foundational to solving problems involving motion and friction. Here, the worker must apply a force equal to the frictional force to maintain constant velocity. This scenario involves calculating several forces, especially focusing on understanding how they relate to each other.

To determine the applied force needed, we start by calculating the normal force, which is the vertical support force exerted by a surface.
  • The normal force equals the gravitational force when there is no vertical movement.
  • Gravitational force is given by \( F_{gravity} = m \cdot g \)
  • The force of kinetic friction then gets calculated using this normal force.
The applied force, in cases of constant velocity, will equal this calculated frictional force, allowing smooth motion without acceleration.
Net Work
Net work done on an object is the total work resulting from all the forces acting on it. In physics, work transfers energy to or from an object, and understanding net work helps determine changes in kinetic energy.

In this exercise, the net work done is found by summing up the work done by all forces.
  • Positive work is done by forces in the direction of motion, like the worker's push.
  • Negative work comes from opposing forces like friction.
  • Perpendicular forces, such as the normal and gravitational forces, contribute zero work.
The result is zero net work because the crate's speed doesn't change — signifying constant velocity. Recognizing zero net work aligns with the principle of no net energy change in constant velocity scenarios.

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Most popular questions from this chapter

A 575 N woman climbs a staircase that rises at \(53^{\circ}\) above the horizontal and is \(4.75 \mathrm{~m}\) long. Her speed is a constant \(45 \mathrm{~cm} / \mathrm{s}\). (a) Is the given weight a reasonable one for an adult woman? (b) How much has the gravitational potential energy increased by her climbing the stairs? (c) How much work has gravity done on her as she climbed the stairs?

On flat ground, a \(70 \mathrm{~kg}\) person requires about \(300 \mathrm{~W}\) of metabolic power to walk at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}(1.4 \mathrm{~m} / \mathrm{s})\) Using the same metabolic power output, that person can bicycle over the same ground at \(15 \mathrm{~km} / \mathrm{h}\). A \(70 \mathrm{~kg}\) person walks at a steady pace of \(5.0 \mathrm{~km} / \mathrm{h}\) on a treadmill at a \(5.0 \%\) grade (that is, the vertical distance covered is \(5.0 \%\) of the horizontal distance covered). If we assume the metabolic power required is equal to that required for walking on a flat surface plus the rate of doing work for the vertical climb, how much power is required? A. \(300 \mathrm{~W}\) B. 315 W C. \(350 \mathrm{~W}\) D. \(370 \mathrm{~W}\)

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When its \(75 \mathrm{~kW}\) ( \(100 \mathrm{hp}\) ) engine is generating full power, a small single-engine airplane with mass \(700 \mathrm{~kg}\) gains altitude at a rate of \(2.5 \mathrm{~m} / \mathrm{s}(150 \mathrm{~m} / \mathrm{min},\) or \(500 \mathrm{ft} / \mathrm{min}) .\) What fraction of the engine power is being used to make the airplane climb? (The remainder is used to overcome the effects of air resistance and of inefficiencies in the propeller and engine.)

Your friend (mass \(65.0 \mathrm{~kg}\) ) is standing on the ice in the middle of a frozen pond. There is very little friction between her feet and the ice, so she is unable to walk. Fortunately, a light rope is tied around her waist and you stand on the bank holding the other end. You pull on the rope for \(3.00 \mathrm{~s}\) and accelerate your friend from rest to a speed of \(6.00 \mathrm{~m} / \mathrm{s}\) while you remain at rest. What is the average power supplied by the force you applied?

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