Top Banner
Introduction to Work
65

Introduction to Work

Jan 19, 2016

Download

Documents

alijah

Introduction to Work. Energy and Work. A body experiences a change in energy when one or more forces do work on it. A body must move under the influence of a force or forces to say work was done. - PowerPoint PPT Presentation
Welcome message from author
This document is posted to help you gain knowledge. Please leave a comment to let me know what you think about it! Share it to your friends and learn new things together.
Transcript
Page 1: Introduction to Work

Introduction to Work

Page 2: Introduction to Work

Energy and Work

A body experiences a change in energy when one or more forces do work on it. A body must move under the influence of a force or forces to say work was done.

A force does positive work on a body when the force and the displacement are at least partially aligned. Maximum positive work is done when a force and a displacement are in exactly the same direction.

If a force causes no displacement, it does zero work.

Forces can do negative work if they are pointed opposite the direction of the displacement.

Page 3: Introduction to Work

Calculating Work in Physics B If a force on an object is at least partially aligned

with the displacement of the object, positive work is done by the force. The amount of work done depends on the magnitude of the force, the magnitude of the displacement, and the degree of alignment.

W= F r cos

r

F

F

Page 4: Introduction to Work

Forces can do positive or negative work. When the load goes

up, gravity does negative work and the crane does positive work.

When the load goes down, gravity does positive work and the crane does negative work.

Ranking Task 1

mg

F

Page 5: Introduction to Work

Units of Work

SI System: Joule (N m)

British System: foot-pound

cgs System: erg (dyne-cm)

Atomic Level: electron-Volt (eV)

Page 6: Introduction to Work

Problem: A droplet of water of mass 50 mg falls at constant speed under the influence of gravity and air resistance. After the drop has fallen 1.0 km, what is the work done by a) gravity and b) air resistance?

Page 7: Introduction to Work

Problem: A sled loaded with bricks has a mass of 20.0 kg. It is pulled at constant speed by a rope inclined at 25o above the horizontal, and it moves a distance of 100 m on a horizontal surface. If the coefficient of kinetic friction between the sled and the ground is 0.40, calculatea)The tension in the rope.

b)The work done by the rope on the sled

c)The work done by friction on the sled.

Page 8: Introduction to Work

Work and a Pulley System

A pulley system, which has at least one pulley attached to the load, can be used to reduce the force necessary to lift a load.

Amount of work done in lifting the load is not changed.

The distance the force is applied over is increased, thus the force is reduced, since W = Fd.

F m

Page 9: Introduction to Work

Work as a “Dot Product”

Page 10: Introduction to Work

Calculating Work a Different Way Work is a scalar resulting from the multiplication of

two vectors. We say work is the “dot product” of force and

displacement. W = F • r

dot product representation W= F r cos

useful if given magnitudes and directions of vectors

W = Fxrx + Fyry + Fzrz

useful if given unit vectors

Page 11: Introduction to Work

The “scalar product” of two vectors is called the “dot product” The “dot product” is one way to multiply two

vectors. (The other way is called the “cross product”.)

Applications of the dot product Work W = F d Power P = F v Magnetic Flux ΦB = B A

The quantities shown above are biggest when the vectors are completely aligned and there is a zero angle between them.

Page 12: Introduction to Work

Why is work a dot product?

s

W = F • rW = F r cos Only the component of force aligned with displacement does work.

F

Page 13: Introduction to Work

•Problem: Vector A has a magnitude of 8.0 and vector B has a magnitude of 12.0. The two vectors make an angle of 40o with each other. Find A•B.

Page 14: Introduction to Work

•Problem: A force F = (5.0i + 6.0j – 2.0k)N acts on an object that undergoes a displacement of r = (4.0i – 9.0j + 3.0k)m. How much work was done on the object by the force?

Page 15: Introduction to Work

• Problem: A force F = (5.0i – 3.0j) N acts upon a body which undergoes a displacement d = (2.0i – j) m. How much work is performed, and what is the angle between the vectors?

Page 16: Introduction to Work

Work by Variable Forces

Page 17: Introduction to Work

Work and Variable Forces

For constant forces W = F • r

For variable forces, you can’t move far until the force changes. The force is only constant over an infinitesimal displacement. dW = F • dr

To calculate work for a larger displacement, you have to take an integral W = dW = F • dr

Page 18: Introduction to Work

Work and variable force

The area under the curve of a graph of force vs displacement gives the work done by the force.

F(x)

xxa xb

W = F(x) dxxa

xb

Page 19: Introduction to Work

• Problem: Determine the work done by the force as the particle moves from x = 2 m to x = 8 m.

F (N)

0

20

40

-20

-40

2 4 6 8 10 12x (m)

Page 20: Introduction to Work

Problem: A force acting on a particle is Fx = (4x – x2)N. Find the work done by the force on the particle when the particle moves along the x-axis from x= 0 to x = 2.0 m.

Page 21: Introduction to Work

• Problem: Derive an expression for the work done by a spring as it is stretched from its equilibrium position

Page 22: Introduction to Work

• Problem: How much work does an applied force do when it stretches a nonlinear spring where the force varies according to the expressions F = (300 N/m) x – (25 N/m2) x2 from its equilibrium length to 20 cm?

Page 23: Introduction to Work

Work Energy Theorem

Page 24: Introduction to Work

Net Work or Total Work

An object can be subject to many forces at the same time, and if the object is moving, the work done by each force can be individually determined.

At the same time one force does positive work on the object, another force may be doing negative work, and yet another force may be doing no work at all.

The net work, or total, work done on the object (Wnet or Wtot) is the scalar sum of the work done on an object by all forces acting upon the object.

Wnet = ΣWi

Page 25: Introduction to Work

The Work-Energy Theorem

Wnet = ΔK When net work due to all forces acting upon an

object is positive, the kinetic energy of the object will increase.

When net work due to all forces acting upon an object is negative, the kinetic energy of the object will decrease.

When there is no net work acting upon an object, the kinetic energy of the object will be unchanged.

(Note this says nothing about the kinetic energy.)

Page 26: Introduction to Work

Kinetic Energy Kinetic energy is one form of mechanical energy,

which is energy we can easily see and characterize. Kinetic energy is due to the motion of an object.

K = ½ m v2

K: Kinetic Energy in Joules. m: mass in kg v: speed in m/s

In vector form, K = ½ m v•v Ranking Tasks 1 2 3 4

Page 27: Introduction to Work

Problem: A net force of 320 N acts over 1.3 m on a 0.4 kg particle moving at 2.0 m/s. What is the speed of this particle after this interaction?

Page 28: Introduction to Work

• Problem: Calculate the kinetic energy change of a 3.0 kg object that changes its velocity from (2.0 i + 2.0 j -1.0 k) m/s to (-1.0 i + 1.0 j -2.0 k) m/s. How much net work done on this object?

Page 29: Introduction to Work

Problem: A force of F1 = (4.0 i + j) N and another of F2 = -4.0 j N act upon a 1 kg object at rest at the origin. What is the speed of the object after it has moved a distance of 3.0 m?

Page 30: Introduction to Work

Power

Power is the rate of which work is done. No matter how fast we get up the stairs, our

work is the same. When we run upstairs, power demands on our

body are high. When we walk upstairs, power demands on our

body are lower. Pave = W / t Pinst = dW/dt P = F • v

Page 31: Introduction to Work

Units of Power

Watt = J/s ft lb / s horsepower

550 ft lb / s 746 Watts

Page 32: Introduction to Work

• Problem: A 1000-kg space probe lifts straight upward off the planet Zombie, which is without an atmosphere, at a constant speed of 3.0 m/s. What is the power expended by the probe’s engines? The acceleration due to gravity of Zombie is ½ that of earth’s.

Page 33: Introduction to Work

• Problem: Develop an expression for the power output of an airplane cruising at constant speed v in level flight. Assume that the aerodynamic drag force is given by FD = bv2.

• By what factor must the power be increased to increase airspeed by 25%?

Page 34: Introduction to Work

How We Buy Energy…

The kilowatt-hour is a commonly used unit by the electrical power company.

Power companies charge you by the kilowatt-hour (kWh), but this not power, it is really energy consumed.

Page 35: Introduction to Work

• Problem: Using what you know about units, calculate how many Joules is in a kilowatt-hour.

Page 36: Introduction to Work

Workday

Page 37: Introduction to Work

Conservative and Non-Conservative Forces

Page 38: Introduction to Work

More about force types

Conservative forces: Work in moving an object is path independent. Work in moving an object along a closed path is zero. Work is directly related to a negative change in potential

energy Ex: gravity, electrostatic, magnetostatic, springs

Non-conservative forces: Work is path dependent. Work along a closed path is NOT zero. Work may be related to a change in mechanical energy, or

thermal energy Ex: friction, drag, magnetodynamic

Page 39: Introduction to Work

Potential Energy

A type of mechanical energy possessed by an object by virtue of its position or configuration.

Represented by the letter U. Examples:

Gravitational potential energy, Ug.

Electrical potential energy , Ue.

Spring potential energy , Us.

The work done by conservative forces is the negative of the potential energy change. W = -ΔU

Page 40: Introduction to Work

Gravitational Potential Energy (Ug) The change in gravitational potential energy is

the negative of the work done by gravitational force on an object when it is moved.

For objects near the earth’s surface, the gravitational pull of the earth is roughly constant, so the force necessary to lift an object at constant velocity is equal to the weight, so we can say

ΔUg = -Wg = mgh Note that this means we have defined the

point at which Ug = 0, which we can do arbitrarily in any given problem close to the earth’s surface.

mg

Fapp

h

Page 41: Introduction to Work

Spring Potential Energy, Us

Springs obey Hooke’s Law. Fs(x) = -kx

Fs is restoring force exerted BY the spring. Ws = Fs(x)dx = -k xdx

Ws is the work done BY the spring. Us = ½ k x2

Unlike gravitational potential energy, we know where the zero potential energy point is for a spring.

Page 42: Introduction to Work

•Problem: Three identical springs (X, Y, and Z) are hung as shown. When a 5.0-kg mass is hung on X, the mass descends 4.0 cm from its initial point. When a 7.0-kg mass is hung on Z, how far does the mass descend? X Y

Z

Page 43: Introduction to Work

• Sample problem: Gravitational potential energy for a body a large distance r from the center of the earth is defined as shown below. Derive this equation from the Universal Law of Gravity.

1 2g

Gm mU

r

• Hint 1: dW = F(r)•dr

• Hint 2: ΔU = -Wc (and gravity is conservative!)

• Hint 3: Ug is zero at infinite separation of the masses.

Page 44: Introduction to Work

Conservation of Mechanical Energy

Page 45: Introduction to Work

System

Boundary

Page 46: Introduction to Work

Law of Conservation of Energy

E = U + K + Eint

= Constant

No mass can enter or leave!No energy can enter or leave!Energy is constant, or conserved!

The system is isolated and boundary allows no exchange with the environment.

Page 47: Introduction to Work

Law of Conservation of Mechanical Energy

E = U + K= Constant

We only allow U and K to interchange.We ignore Eint (thermal energy)

Page 48: Introduction to Work

Law of Conservation of Mechanical Energy E = U + K = C or E = U + K = 0 for gravity

Ug = mghf - mghi

K = ½ mvf2 - ½ mvi

2

(What assumptions are we making here?) for springs

Us = ½ kxf2 - ½ kxi

2 K = ½ mvf

2 - ½ mvi2

(What assumptions are we making here?) Ranking Tasks 1

Page 49: Introduction to Work

h

Pendulum Energy

½mv12 + mgh1 = ½mv2

2 + mgh2 For any points two points in the pendulum’s swing

Page 50: Introduction to Work

Spring Energy

m

m -x

mx

0

½ kx12 + ½ mv1

2

= ½ kx22 + ½

mv22

For any two points in a spring’s oscillation

Page 51: Introduction to Work

• Problem: A single conservative force of F = (3i + 5j) N acts on a 4.0 kg particle. Calculate the work done if the particle if the moves from the origin to r = (2i - 3j) m. Does the result depend on path?

• What is the speed of the particle at r if the speed at the origin was 4.0 m/s?

• What is the change in potential energy of the system?

Page 52: Introduction to Work

•Sample Problem: A bead slides on the loop-the-loop shown. If it is released from height h = 3.5 R, what is the speed at point A? How great is the normal force at A if the mass is 5.0 g?

Page 53: Introduction to Work

Non-conservative Forces and Conservation of Energy

Page 54: Introduction to Work

Non-conservative forces

Non-conservative forces change the mechanical energy of a system.

Examples: friction and drag Wtot = Wnc + Wc = K

Wnc = K – Wc

Wnc = K + U

Page 55: Introduction to Work

Sample Problem: A 2,000 kg car starts from rest and coasts down from the top of a 5.00 m long driveway that is sloped at an angel of 20o with the horizontal. If an average friction force of 4,000 N impedes the motion of the car, find the speed of the car at the bottom of the driveway.

Page 56: Introduction to Work

• Problem: A parachutist of mass 50 kg jumps out of a hot air balloon 1,000 meters above the ground and lands on the ground with a speed of 5.00 m/s. How much energy was lost to friction during the descent?

Page 57: Introduction to Work

Conservation of Energy Lab

Page 58: Introduction to Work

Force and Potential Energy

In order to discuss the relationships between potential energy and force, we need to review a couple of relationships.

Wc = Fx (if force is constant)

Wc = Fdx = - dU = -U (if force varies)

Fdx = - dU Fdx = -dU F = -dU/dx

Page 59: Introduction to Work

Stable Equilibrium – 1st and 2nd Derivatives

U

x

Page 60: Introduction to Work

Unstable Equilibrium – 1st and 2nd Derivatives

U

x

Page 61: Introduction to Work

Neutral Equilibrium – 1st and 2nd Derivatives

U

x

Page 62: Introduction to Work

More on Potential Energy and Force In multiple dimensions, you can take

derivatives of each dimension separately. F(r) = -dU(r)/dr

Fx = -U/ x

Fy = -U/ y

Fz = -U/ z

F = Fxi + Fyj + Fzk

Page 63: Introduction to Work

Molecular potential energy diagrams

R

U

Page 64: Introduction to Work

• Problem: The potential energy of a two-particle system separated by a distance r is given by U(r) = A/r, where A is a constant. Find the radial force F that each particle exerts on the other.

Page 65: Introduction to Work

• Problem: A potential energy function for a two-dimensional force is of the form U = 3x3y – 7x. Find the force acting at a point (x,y).