/*! 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 3 A factory worker pushes a \(30.0... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 3

A factory worker pushes 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 this force? (c) How much work is done on the crate by friction? (d) How much work is done on the crate by the normal force? By gravity? (e) What is the total work done on the crate?

Short Answer

Expert verified
The magnitude of force that the worker must apply is 73.5N. The work done on the crate by this force is 330.75J. The work done on the crate by friction is -330.75J. As the normal force and gravity are perpendicular to displacement, they both do no work on the crate, hence 0J. The total work done on the crate is 0J.

Step by step solution

01

Calculate Force applied by the Worker

We first calculate the normal force which in this case is equal to the weight of the crate and then use it to find the force applied by the worker. It's given by \(F_{n} = m*g = 30.0kg * 9.8m/s^2 = 294N\). Then, \(F_{w} = \mu_{k}*F_{n} => 0.25 * 294N = 73.5N\). So, the worker must apply a force of 73.5N to maintain the constant velocity.
02

Calculate Work Done by the Worker

The work done by a force is given by the formula \(\ W = F * d * cos(\theta)\). Since the force is applied in the same direction as the movement \(\theta = 0\), therefore, cos(\theta) = 1. So, \(W = F_{w} * d = 73.5N * 4.5m = 330.75J\). The work done by the worker is therefore 330.75 Joules.
03

Calculate Work done by Friction

The work done by friction would be in negative because it acts in the direction opposite to the displacement. \(W_{f} = -F_{f} * d = -73.5N * 4.5m = -330.75J\). Thus, the work done by friction is -330.75 Joules.
04

Calculate work done by Normal force and Gravity

Both the normal force and gravity act perpendicular to the direction of movement, thus they do no work on the crate. So, \(W_{n}=W_{g}=0J\).
05

Calculate total work done

The total work done on the crate is the sum of the work done by the worker, the work done by friction, and the work done by gravity and the normal force. \( W_{t}= W_{w} + W_{f} + W_{n} +W_{g} = 330.75J - 330.75J + 0J + 0J = 0J\). Thus, the total work done on the crate is 0 Joules.

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 a surface, they encounter a resisting force known as kinetic friction. This force acts in the opposite direction to the motion. In our problem, the worker is pushing a 30 kg crate across a floor. The coefficient of kinetic friction, represented as \(\mu_k\), is given as 0.25. This value is essential for calculating how much force one needs to keep the crate moving at a constant velocity. The kinetic friction is calculated using the formula:
  • \(F_{f} = \mu_k \times F_n\)
Here, \(F_f\) is the force of friction, and \(F_n\) is the normal force. Since the crate is moving along a level surface, the normal force equates to the crate's weight. Thus, to maintain a constant velocity, the worker must counteract this frictional force by applying an equal and opposite force.
Force Calculation
To find out the magnitude of the force the worker needs to apply, we first calculate the normal force. The normal force is approximately the force exerted by the floor to support the weight of the crate. It can be calculated using the formula:
  • \(F_n = m \times g\)
Where \(m\) is the mass of the crate (30 kg), and \(g\) is the acceleration due to gravity (approximated to 9.8 m/s虏). This yields \(F_n = 294 \text{ N}\).
The frictional force is then:
  • \(F_f = \mu_k \times F_n = 0.25 \times 294 \text{ N} = 73.5 \text{ N}\)
Thus, the worker must apply a force of 73.5 N to keep the crate moving steadily.
Work Done by Forces
Work in physics is defined as the force applied to an object multiplied by the distance over which the force is applied. The formula to calculate work (\(W\)) is:
  • \(W = F \times d \times \cos(\theta)\)
Since the worker's force is in the same direction as the movement, \(\theta = 0\), making \(\cos(\theta) = 1\). Thus, the work done by the worker is:
\(W_w = 73.5 \text{ N} \times 4.5 \text{ m} = 330.75 \text{ J}\).
The work done by friction opposes this, given by:
\(W_f = -73.5 \text{ N} \times 4.5 \text{ m} = -330.75 \text{ J}\).
Since friction acts in the opposite direction, its work is negative, cancelling out the work done by the applied force.
Normal Force
The normal force is the perpendicular force exerted by a surface to support the weight of an object resting upon it. In this scenario, the normal force corresponds to the weight of the crate, preventing it from sinking into the floor. Neither the normal force nor gravity does work in this problem since they are both perpendicular to the crate's horizontal motion. Work is associated only with the component of force in the direction of movement, and in this case:
  • \(W_n = 0 \text{ J}\)
  • \(W_g = 0 \text{ J}\)
Both these components contribute zero to the total work done on the crate, which ensures the problem鈥檚 net total work remains zero.鈥�

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

Proton Bombardment. A proton with mass \(1.67 \times 10^{-27} \mathrm{~kg}\) is propelled at an initial speed of \(3.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a uranium nucleus \(5.00 \mathrm{~m}\) away. The proton is repelled by the uranium nucleus with a force of magnitude \(F=\alpha / x^{2},\) where \(x\) is the separation between the two objects and \(\alpha=2.12 \times 10^{-26} \mathrm{~N} \cdot \mathrm{m}^{2}\). Assume that the uranium nucleus remains at rest. (a) What is the speed of the proton when it is \(8.00 \times 10^{-10} \mathrm{~m}\) from the uranium nucleus? (b) As the proton approaches the uranium nucleus, the repulsive force slows down the proton until it comes momentarily to rest, after which the proton moves away from the uranium nucleus. How close to the uranium nucleus does the proton get? (c) What is the speed of the proton when it is again \(5.00 \mathrm{~m}\) away from the uranium nucleus?

How many joules of energy does a 100 watt light bulb use per hour? How fast would a \(70 \mathrm{~kg}\) person have to run to have that amount of kinetic energy?

A \(5.00 \mathrm{~kg}\) block is moving at \(v_{0}=6.00 \mathrm{~m} / \mathrm{s}\) along a frictionless, horizontal surface toward a spring with force constant \(k=500 \mathrm{~N} / \mathrm{m}\) that is attached to a wall (Fig. P6.79). The spring has negligible mass. (a) Find the maximum distance the spring will be compressed. (b) If the spring is to compress by no more than \(0.150 \mathrm{~m},\) what should be the maximum value of \(v_{n} ?\)

Stopping Distance. A car is traveling on a level road with speed \(v_{0}\) at the instant when the brakes lock, so that the tires slide rather than roll. (a) Use the work-energy theorem to calculate the minimum stopping distance of the car in terms of \(v_{0}, g,\) and the coefficient of kinetic friction \(\mu_{\mathrm{k}}\) between the tires and the road. ( b) By what factor would the minimum stopping distance change if (i) the coefficient of kinetic friction were doubled, or (ii) the initial speed were doubled, or (iii) both the coefficient of kinetic friction and the initial speed were doubled?

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 that 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

Electromagnetism

Read Explanation

Medical Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Quantum Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Fundamentals of 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.