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Chapter 6: Problem 3

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 \(-2.55 \times 10^{6} \, \text{J}\) and it's negative because the force opposes motion.

Step by step solution

01

Identify the Given Values

We are given a retarding force of \(3.0 \times 10^{3} \, \mathrm{N}\) and a distance of \(850 \, \mathrm{m}\). The question asks for the work done by the force applied to the truck.
02

Understand the Concept of Work

Work done by a force is calculated using the formula \( W = F \cdot d \cdot \cos(\theta) \), where \( F \) is the force, \( d \) is the distance over which the force is applied, and \( \theta \) is the angle between the force and the direction of motion.
03

Determine the Angle between Force and Motion

Since the force is retarding, it acts in the opposite direction to the motion of the truck. Therefore, \( \theta = 180^{\circ} \). The cosine of \(180^{\circ}\) is \(-1\).
04

Plug Values into the Work Formula

Substitute the values into the work formula: \[ W = 3.0 \times 10^{3} \, \text{N} \times 850 \, \text{m} \times \cos(180^{\circ}) \]\[ W = 3.0 \times 10^{3} \, \text{N} \times 850 \, \text{m} \times (-1) \]\[ W = -2.55 \times 10^{6} \, \text{J} \]
05

Determine Sign and Explanation of Work

The work done is \(-2.55 \times 10^{6} \, \text{J}\). The negative sign indicates that the work is done against the direction of motion of the truck, which is typical for a retarding or braking force.

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Key Concepts

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

Understanding Retarding Force
When a truck slows down, a retarding force is at work. This is a force that opposes the motion of the object. In our scenario, the brakes apply a retarding force of \(3.0 \times 10^3 \text{ N}\), which means they exert 3000 Newtons against the truck's motion.
Consider retarding forces as the brakes that halt your moving skateboard. They don't just stop you instantly but apply a steady push back against your forward motion. This deceleration requires energy, which is where the concept of work comes into play. The force does work on the truck by losing kinetic energy and converting it into other forms, like heat due to friction.
This force is critical in many practical applications such as braking systems, where controlling motion safely is vital.
The Angle of Force Application
In physics, the angle of force application is crucial to calculating work done. The work formula is \( W = F \cdot d \cdot \cos(\theta) \). Here, \(\theta\) is the angle between the force applied and the direction of motion.
For the truck, the retarding force and direction of motion are opposite, creating an angle of \(180^{\circ}\). This opposite direction results in a cosine of \(-1\).
Notice how the angle influences the work done by the force:
  • At \(0^{\circ}\), force and motion align, maximizing work done.
  • At \(90^{\circ}\), force is perpendicular to motion, resulting in no work done.
  • At \(180^{\circ}\), force counteracts motion, causing negative work.
Understanding these angles helps us decipher why certain actions slow or speed up objects.
Exploring Negative Work Direction
The concept of negative work direction is key when dealing with forces that oppose motion. In this exercise, the negative sign in the work calculation \(-2.55 \times 10^6 \text{ J}\) signifies that the retarding force is working against the truck's motion.
This negative work doesn't mean energy loss in the traditional sense but a transfer from kinetic to other energy forms due to opposition. When brakes are used, energy is absorbed to decelerate, performing work opposite to the truck's motion.
Understanding negative work helps reveal how mechanisms like brakes affect motion:
  • Explain why cars slow down efficiently.
  • Understand energy conversion during deceleration.
  • Analyze effects in everyday systems like walking down a hill.
It emphasizes the role that oppositional forces play in energy transformation during motion.

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Most popular questions from this chapter

The drawing shows a skateboarder moving at \(5.4 \mathrm{m} / \mathrm{s}\) along a horizontal section of a track that is slanted upward by \(48^{\circ}\) above the horizontal at its end, which is \(0.40 \mathrm{m}\) above the ground. When she leaves the track, she follows the characteristic path of projectile motion. Ignoring friction and air resistance, find the maximum height \(H\) to which she rises above the end of the track.

You are moving into an apartment and take the elevator to the 6 th floor. Suppose your weight is \(685 \mathrm{N}\) and that of your belongings is \(915 \mathrm{N}\). (a) Determine the work done by the elevator in lifting you and your belongings up to the 6 th floor \((15.2 \mathrm{m})\) at a constant velocity. (b) How much work does the elevator do on you alone (without belongings) on the downward trip, which is also made at a constant velocity?

A slingshot fires a pebble from the top of a building at a speed of \(14.0 \mathrm{m} / \mathrm{s} .\) The building is \(31.0 \mathrm{m}\) tall. Ignoring air resistance, find the speed with which the pebble strikes the ground when the pebble is fired (a) horizontally, (b) vertically straight up, and (c) vertically straight down.

A water-skier is being pulled by a tow rope attached to a boat. As the driver pushes the throttle forward, the skier accelerates. A \(70.3-\mathrm{kg}\) water-skier has an initial speed of \(6.10 \mathrm{m} / \mathrm{s}\). Later, the speed increases to \(11.3 \mathrm{m} / \mathrm{s} .\) Determine the work done by the net external force acting on the skier.

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?
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