/*! 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 18 In 1975 the roof of Montreal's V... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 18

In 1975 the roof of Montreal's Velodrome, with a weight of \(360 \mathrm{kN},\) was lifted by \(10 \mathrm{~cm}\) so that it could be centered. How much work was done on the roof by the forces making the lift? (b) In 1960 a Tampa, Florida, mother reportedly raised one end of a car that had fallen onto her son when a jack failed. If her panic lift effectively raised \(4000 \mathrm{~N}\) (about \(\frac{1}{4}\) of the car's weight) by \(5.0 \mathrm{~cm},\) how much work did her force do on the car?

Short Answer

Expert verified
The work done on the roof is 36,000 J; the work done by the mother is 200 J.

Step by step solution

01

Understanding Work Done

Work is defined as the product of force and distance when a force causes a displacement. It can be calculated with the formula: \[ W = F \times d \]where \( W \) is the work done, \( F \) is the force applied, and \( d \) is the displacement in the direction of the force.
02

Calculate Work on the Velodrome Roof

The weight of the Velodrome roof is given as \( 360 \mathrm{kN} = 360,000 \mathrm{~N} \). The roof was lifted \( 10 \mathrm{~cm} = 0.1 \mathrm{~m} \). Using the work formula, \[ W = 360,000 \mathrm{~N} \times 0.1 \mathrm{~m} = 36,000 \mathrm{~J} \]Thus, the work done on the roof is \( 36,000 \mathrm{~J} \).
03

Calculate Work Done by the Mother

The force exerted by the mother is \( 4000 \mathrm{~N} \) and the car was lifted \( 5.0 \mathrm{~cm} = 0.05 \mathrm{~m} \). Using the work formula, \[ W = 4000 \mathrm{~N} \times 0.05 \mathrm{~m} = 200 \mathrm{~J} \]Thus, the work done by the mother on the car is \( 200 \mathrm{~J} \).

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.

Force
In physics, force is a key concept related to motion and energy. It is defined as an interaction that changes the motion of an object when there is a physical contact or at a distance, such as gravity. Force is measured in newtons (N).
  • When you push, pull, or lift an object, you are applying force.
  • Force can cause an object to start moving, stop moving, or change directions.
  • The amount of force you apply can affect how much work is done.

In the exercises provided, two different forces are mentioned: the force required to lift the roof of the Velodrome, which is quite large at 360,000 N, and the force applied by a mother lifting part of a car at 4000 N. The differences in these forces are determined by the weight and the situation involved. Understanding how force works, helps in calculating the work done.
Displacement
Displacement refers to how far an object has moved from its original position in a particular direction. This is different from distance traveled, as displacement considers direction.
  • It is measured in meters (m).
  • It is a vector quantity, meaning it has both magnitude and direction.

In our examples, the displacement for the Velodrome roof was 0.1 meters, and for the car lift, it was 0.05 meters. Even though the distances might seem small, they are significant enough to calculate the energy needed in these scenarios. Realizing the role of displacement helps in understanding how force leads to work.
Work Formula
The work formula is essential in determining how much work is done when a force causes an object to move. The basic formula for work is: \[ W = F \times d \]where:
  • \( W \) is the work done (measured in joules, J),
  • \( F \) is the force applied (measured in newtons, N),
  • \( d \) is the displacement (measured in meters, m) in the direction of the force.

The concept of work in physics describes an energy transfer when an object is moved over a distance by an external force. For the Velodrome roof scenario, applying the work formula with the known force and displacement gives a work done of 36,000 joules. Similarly, for the mother's heroic act, it results in work done of 200 joules. Understanding this formula is crucial for calculating energy in physics.
Energy in Physics
Energy is the capacity to do work, and it can exist in various forms, such as kinetic or potential energy. In physics, energy is closely tied to work because doing work on an object transfers energy to it.
  • Energy can be converted from one form to another, but the total energy in a closed system remains constant (conservation of energy).
  • Work done on an object results in a change in energy; for example, lifting objects increases their gravitational potential energy.

In the Velodrome and Tampa situations, lifting the roof or raising the car end required energy. This energy was transferred from the force exerted (like muscular energy from the mother) to the objects being moved. Knowing how energy works in conjunction with force and work helps us understand the underlying principles of dynamics and mechanics in physics.

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

61 How much work is done by a force \(\vec{F}=(2 x \mathrm{~N}) \hat{\mathrm{i}}+(3 \mathrm{~N}) \hat{\mathrm{j}},\) with \(x\) in meters, that moves a particle from a position \(\vec{r}_{i}=\) \((2 \mathrm{~m}) \hat{\mathrm{i}}+(3 \mathrm{~m}) \hat{\mathrm{j}}\) to a position \(\vec{r}_{f}=-(4 \mathrm{~m}) \hat{\mathrm{i}}-(3 \mathrm{~m}) \hat{\mathrm{j}} ?\)

If a Saturn \(\mathrm{V}\) rocket with an Apollo spacecraft attached had a combined mass of \(2.9 \times 10^{5} \mathrm{~kg}\) and reached a speed of \(11.2 \mathrm{~km} / \mathrm{s}\), how much kinetic energy would it then have?

The loaded cab of an elevator has a mass of \(3.0 \times 10^{3} \mathrm{~kg}\) and moves \(210 \mathrm{~m}\) up the shaft in \(23 \mathrm{~s}\) at constant speed. At what average rate does the force from the cable do work on the cab?

A particle moves along a straight path through displacement \(\vec{d}=(8 \mathrm{~m}) \hat{\mathrm{i}}+c \hat{\mathrm{j}}\) while force \(\vec{F}=(2 \mathrm{~N}) \hat{\mathrm{i}}-(4 \mathrm{~N}) \hat{\mathrm{j}}\) acts on it. (Other forces also act on the particle.) What is the value of \(c\) if the work done by \(\vec{F}\) on the particle is (a) zero, (b) positive, and (c) negative?

A force \(\vec{F}\) in the positive direction of an \(x\) axis acts on an object moving along the axis. If the magnitude of the force is \(F=10 e^{-x / 2.0} \mathrm{~N}\), with \(x\) in meters, find the work done by \(\vec{F}\) as the object moves from \(x=0\) to \(x=2.0 \mathrm{~m}\) by \((\) a \()\) plotting \(F(x)\) and estimating the area under the curve and (b) integrating to find the work analytically.
See all solutions

Recommended explanations on Physics Textbooks

Translational Dynamics

Read Explanation

Atoms and Radioactivity

Read Explanation

Linear Momentum

Read Explanation

Energy Physics

Read Explanation

Thermodynamics

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.