Chapter 6: Work - PowerPoint PPT Presentation

Loading...

PPT – Chapter 6: Work PowerPoint presentation | free to download - id: 201730-ZDc1Z



Loading


The Adobe Flash plugin is needed to view this content

Get the plugin now

View by Category
About This Presentation
Title:

Chapter 6: Work

Description:

Everyday meanings might not coincide with the PHYSICS meaning! ... 'Kinetic' Greek word for motion. An object in motion has the ability to do work. ... – PowerPoint PPT presentation

Number of Views:25
Avg rating:3.0/5.0
Slides: 27
Provided by: Charles4
Learn more at: http://www.phys.ttu.edu
Category:
Tags: chapter | greek | meanings | word | work

less

Write a Comment
User Comments (0)
Transcript and Presenter's Notes

Title: Chapter 6: Work


1
Chapter 6 Work Energy
2
  • THE COURSE THEME is
  • NEWTONS LAWS OF MOTION!
  • Chs. 4, 5 Motion analysis with forces.
  • NOW (Ch. 6) An alternative analysis using the
    concepts of Work Energy.
  • Easier? My opinion is yes!
  • Conservation of Energy NOT a new law!
  • Well see that this is just Newtons Laws of
    Motion re-formulated or re-expressed (translated)
    from Force Language to Energy Language.

3
  • Weve expressed Newtons Laws of Motion using the
    concepts of position, displacement, velocity,
    acceleration force.
  • Newtons Laws with Forces Quite general
    (macroscopic objects). In principle, could be
    used to solve any dynamics problem, But, often,
    they are very difficult to apply, especially to
    very complicated systems. So, alternate
    formulations have been developed. Often easier to
    apply.
  • Newtons 2nd Law
  • Often we may not even know all of the forces.
  • One alternate approach uses Energy instead of
    Force as the most basic physical quantity.
  • Newtons Laws in a different language (Energy).
    Before we discuss these, we need to learn some
    vocabulary in Energy Language .
  • Energy A very common term in everyday usage.
    Everyday meanings might not coincide with the
    PHYSICS meaning!
  • Every physical process involves energy or energy
    transfer or transformations. Energy in physics
    can be somewhat abstract.

4
  • So far, weve expressed Newtons Laws of Motion
    in terms of forces weve considered the
    dynamical properties of a particle by talking
    about various particle properties.
  • Now, well take a different approach talk about
    Systems System Properties.
  • System A small portion of the universe which we
    focus on in a given problem. What the system is
    depends on the problem.
  • A System may be, for example
  • A single particle.
  • A collection of particles.
  • A region of space.
  • May vary in size shape, depending on the
    problem
  • In addition to a System, we also talk about the
    System environment. The System interacts with
    environment at its boundaries.

5
Sect. 6-1 Work Done by Constant Force
  • Work Precisely defined in physics. Describes
    what is accomplished by a force in moving an
    object through a distance.

For an object moving under the influence of a
Constant Force, the work done (W) ? the product
of the magnitude of the displacement (d) ? the
component of force parallel to the displacement
(F). W ? Fd Fd
cos?
d
6
Work Done by a Constant Force
Work W ? Fd Fd cos? For a CONSTANT force!
7
  • W Fd Fd cos?
  • Consider a simple special case when F d are
    parallel
  • ? 0, cos? 1
  • ? W Fd
  • Example d 50 m, F 30 N
  • W (30N)(50m) 1500 N m
  • Work units Newton - meter Joule
  • 1 N m 1 Joule 1 J

8
  • W Fd Fd cos?
  • Can exert a force do no work!
  • Could have d 0 ? W 0
  • Could have F ? d
  • ? ? 90º, cos? 0
  • ? W 0
  • Example, walking at constant v
  • with a grocery bag

9
Example 6-1
  • W Fd Fd cos?
  • m 50 kg, FP 100 N, Ffr 50 N, ? 37º

10
  • Solving Work Problems
  • Sketch a free-body diagram.
  • Choose a coordinate system.
  • Apply Newtons Laws to determine any unknown
    forces.
  • Find the work done by a specific force.
  • Find the net work by either
  • a. Find the net force then find the work it
    does, or
  • b. Find the work done by each force add.

11
  • W Fd Fd cos?

A Typical Problem An object
displaced by force F on a frictionless,
horizontal surface. The free body diagram is
shown. ? The normal force FN
weight mg do no work in the process, since both
are perpendicular to the displacement. Angles
for forces Normal force ? 90, cos?
0 Weight ? 270 (or - 90), cos? 0
FN
d
12
Ex. 6-2 The work on a backpack
(a) Calculate the work a hiker must do on a
backpack of mass m 15 kg in order to
carry it up a hill of height h 10 m, as
shown. (b) Calculate the work done by gravity on
the backpack. (c) Calculate the net work done on
the backpack. For simplicity, assume that
the motion is smooth at constant
velocity (zero acceleration).
For the hiker, ?Fy 0 FH - mg
? FH mg WH FHdcos? FHh
13
Conceptual Ex. 6-3 Does the Earth do work on
the Moon?
The Moon revolves around the Earth in a nearly
circular orbit, with approximately constant
tangential speed, kept there by the gravitational
force exerted by the Earth. Does gravity do (a)
positive work (b) negative work, or (c) no work
at all on the Moon?
14
Example
The force shown has magnitude FP 20 N makes
an angle ? 30 to the ground. Calculate the
work done by this force when the wagon is dragged
a displacement d 100 m along the ground.
15
Sect. 6-3 Kinetic Energy Work-Energy Principle
16
  • Energy Traditionally defined as the ability to
    do work. We now know that not all forces are able
    to do work however, we are dealing in these
    chapters with mechanical energy, which does
    follow this definition.
  • Kinetic Energy ? The energy of motion
  • Kinetic ? Greek word for motion
  • An object in motion has the ability to do work.

17
  • Consider an object moving in straight line. It
    starts at speed v1. Due to the presence of a net
    force Fnet, ( ?F), it accelerates (uniformly) to
    speed v2, over a distance d.
  • Newtons 2nd Law Fnet ma (1)
  • 1d motion, constant a
  • ? (v2)2 (v1)2 2ad
  • ? a (v2)2 - (v1)2/(2d) (2)
  • Work done Wnet Fnet d (3)
  • Combine (1), (2), (3)

18
  • Fnet ma (1)
  • a (v2)2 - (v1)2/(2d) (2)
  • Wnet Fnet d (3)
  • Combine (1), (2) (3)
  • ? Wnet mad md (v2)2 - (v1)2/(2d)
  • OR
  • Wnet (½)m(v2)2 (½)m(v1)2

19
  • Summary The net work done by a constant force in
    accelerating an object of mass m from v1 to v2
    is

  • ? ?KE
  • DEFINITION Kinetic Energy (KE)
  • (for translational motion Kinetic
    motion)
  • (units
    are Joules, J)
  • Weve shown The WORK-ENERGY PRINCIPLE
  • Wnet ?KE (? change in)

Weve shown this for a 1d constant force.
However, it is valid in general!
20
  • Net work on an object Change in KE.
  • Wnet ?KE (I)
  • ? The Work-Energy Principle
  • Note Wnet work done by the net (total) force.
  • Wnet is a scalar can be positive or negative
    (because ?KE can be both -). If the net work
    is positive, the kinetic energy KE increases. If
    the net work is negative, the kinetic energy KE
    decreases.
  • Units are Joules for both work kinetic
    energy.
  • Note (I) is Newtons 2nd Law in
  • Work Energy language!

21
  • A moving hammer can do work on a nail!
  • For the hammer
  • Wh ?KEh -Fd
  • 0 (½)mh(vh)2
  • For the nail
  • Wn ?KEn Fd
  • (½)mn(vn)2 - 0

22
Example 6-4 Kinetic energy
work done on a baseball
A baseball, mass m 145 g (0.145 kg) is thrown
so that it acquires a speed v 25 m/s. a.
What is its kinetic energy? b. What was the
net work done on the ball to make it
reach this speed, starting from rest?
23
Ex. 6-5 Work on a car to increase its kinetic
energy
Calculate the net work required to accelerate a
car, mass m 1000-kg car from v1 20 m/s to v2
30 m/s.
24
Conceptual Example 6-6 Work to stop a car
A car traveling at speed v1 60 km/h can brake
to a stop within a distance d 20 m. If the car
is going twice as fast, 120 km/h, what is its
stopping distance? Assume that the maximum
braking force is approximately independent of
speed.
25
Wnet Fd cos (180º) -Fd (from the
definition of work) Wnet ?KE (½)m(v2)2
(½)m(v1)2 (Work-Energy Principle) but, (v2)2 0
(the car has stopped) so -Fd ?KE 0 -
(½)m(v1)2 or d ? (v1)2 So the stopping
distance is proportional to the square of the
initial speed! If
the initial speed is doubled, the stopping
distance quadruples! Note KE ? (½)mv2 ? 0 Must
be positive, since m v2 are always positive
(real v).
26

Example
A block, mass m 6 kg, is pulled from rest (v0
0) to the right by a constant horizontal force
F 12 N. After it has been pulled for ?x 3
m, find its final speed v. Work-Energy
Principle Wnet ?KE ? (½)m(v)2 - m(v)2
(1) If F 12 N is the only horizontal force, we
have Wnet F?x (2) Combine
(1) (2) F?x (½)m(v)2 - 0
Solve for v (v)2 2?x/m v
2?x/m½ 3.5 m/s
l
l
About PowerShow.com