/*! 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 The brakes of a truck cause it t... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The brakes of a truck cause it to slow down by applying a retarding force of \(3.0 \times 10^{3} \mathrm{N}\) to the truck over a distance of \(850 \mathrm{m} .\) What is the work done by this force on the truck? Is the work positive or negative? Why?

Short Answer

Expert verified
The work done is \(-2.55 \times 10^{6} \text{ J}\). It is negative because the force opposes the truck's motion.

Step by step solution

01

Identify known values

Start by identifying the values given in the problem. We have a retarding force of \( F = 3.0 \times 10^{3} \, \text{N} \) and a distance \( d = 850 \, \text{m} \). These are the key values we need to use in the work formula.
02

Recall the formula for work

The work done by a force is given by the formula \( W = F \times d \times \cos(\theta) \), where \( \theta \) is the angle between the force and the direction of motion. In this case, the force is directly opposite to the direction of motion, making \( \theta = 180^\circ \).
03

Calculate cosine of the angle

Since the angle \( \theta \) is \( 180^\circ \), the cosine of \( 180^\circ \) is \( -1 \). This indicates that the force applied is in the opposite direction to the motion.
04

Substitute values into the work formula

Insert the known values into the work formula: \( W = 3.0 \times 10^{3} \, \text{N} \times 850 \, \text{m} \times (-1) \). Calculate the work: \( W = -2.55 \times 10^{6} \, \text{J} \).
05

Interpret the result

The negative sign in the result indicates that the work done by the force on the truck is negative. This tells us that the force is acting in the opposite direction to the truck’s motion, thus decreasing its kinetic energy.

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 91Ó°ÊÓ!

Key Concepts

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

Retarding Force
The retarding force is a type of force that acts opposite to the direction of an object's motion. It reduces the speed of the object, essentially slowing it down. In our exercise, the brakes of the truck apply this retarding force to stop the truck from moving faster. The key idea here is that a retarding force always does work on an object, but not in the way that other forces might. Instead of increasing energy, it takes energy away from the object's motion.
Retarding forces are commonly seen in everyday situations:
  • Brakes on cars and bicycles, which slow them down or bring them to a complete stop.
  • Frictional forces that oppose natural motion, like a sliding block on a rough surface.
Understanding retarding force is crucial in applications where control over speed is necessary. It is vital in ensuring safety by decreasing kinetic energy, such as stopping a vehicle within a safe distance.
Distance
Distance in physics refers to the measure of how much ground an object has covered during its motion. It's a scalar quantity, meaning it only has magnitude and not direction. In the original exercise, the distance is specified as 850 meters, which is the path over which the retarding force is applied. Knowing the distance is essential for calculating work done, since the formula for work is dependent on how far the object travels under the influence of a force.
When calculating work done by a force:
  • Distance is directly proportional to work. As distance increases, the work done also increases if the force remains constant.
  • If no distance is covered, no work is done.
Understanding the concept of distance helps in analyzing the effectiveness of a force applied to move or stop an object over a certain path.
Angle of Force
When evaluating work, the angle of force, symbolized as \( \theta \), represents the angle between the direction of the force applied and the direction of the object’s motion. It's crucial because it determines how much of the force actually contributes to the work performed in the direction of motion. In the exercise, the angle is \( 180^\circ \), indicating the force is directly opposite to the motion.
This opposition is significant:
  • When \( \theta = 0^\circ \), the force is in the direction of motion, resulting in maximum positive work.
  • When \( \theta = 90^\circ \), no work is done because the force is perpendicular to the motion.
  • When \( \theta = 180^\circ \), like in our exercise, the work done is negative. This is because the force decreases the object's kinetic energy, slowing it down.
Understanding how the angle of force influences work allows students to grasp why certain forces result in positive or negative work depending on their alignment with the motion.

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

A 0.60-kg basketball is dropped out of a window that is 6.1 m above the ground. The ball is caught by a person whose hands are \(1.5 \mathrm{m}\) above the ground. (a) How much work is done on the ball by its weight? What is the gravitational potential energy of the basketball, relative to the ground, when it is (b) released and (c) caught? (d) How is the change \(\left(\mathrm{PE}_{\mathrm{f}}-\mathrm{PE}_{0}\right)\) in the ball's gravitational potential energy related to the work done by its weight?

In the sport of skeleton a participant jumps onto a sled (known as a skeleton) and proceeds to slide down an icy track, belly down and head first. In the 2010 Winter Olympics, the track had sixteen turns and dropped \(126 \mathrm{m}\) in elevation from top to bottom. (a) In the absence of nonconservative forces, such as friction and air resistance, what would be the speed of a rider at the bottom of the track? Assume that the speed at the beginning of the run is relatively small and can be ignored. (b) In reality, the gold-medal winner (Canadian Jon Montgomery) reached the bottom in one heat with a speed of \(40.5 \mathrm{m} / \mathrm{s}\) (about \(91 \mathrm{mi} / \mathrm{h}\) ). How much work was done on him and his sled (assuming a total mass of \(118 \mathrm{kg}\) ) by nonconservative forces during this heat?

A \(16-\mathrm{kg}\) sled is being pulled along the horizontal snow-covered ground by a horizontal force of \(24 \mathrm{N}\). Starting from rest, the sled attains a speed of \(2.0 \mathrm{m} / \mathrm{s}\) in \(8.0 \mathrm{m} .\) Find the coefficient of kinetic friction between the runners of the sled and the snow.

A golf club strikes a 0.045-kg golf ball in order to launch it from the tee. For simplicity, assume that the average net force applied to the ball acts parallel to the ball's motion, has a magnitude of \(6800 \mathrm{N},\) and is in contact with the ball for a distance of \(0.010 \mathrm{m}\). With what speed does the ball leave the club?

A person is making homemade ice cream. She exerts a force of magnitude \(22 \mathrm{N}\) on the free end of the crank handle on the ice-cream maker, and this end moves on a circular path of radius \(0.28 \mathrm{m}\). The force is always applied parallel to the motion of the handle. If the handle is turned once every \(1.3 \mathrm{s},\) what is the average power being expended?

See all solutions

Recommended explanations on Physics Textbooks

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.