Post on 26-Mar-2015
Introduction to Work
Thursday, Jan 6, 2011! Happy New Year.
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.
Calculating Work – Step 1, constant F 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
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
Units of Work
SI System: Joule (N m)
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?
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.
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
Work as a “Dot Product”
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
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.
Why is work a dot product?
s
W = F • rW = F r cos Only the component of force aligned with displacement does work.
F
•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.
•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?
• 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?
Homework 1/6/11
• Page 160-163: 18, 19, 21, 51, 56
Work by Variable Forces
Sometimes, forces are not constant. Examples, spring forces, air resistant forces. How do we calculate work in these situations?
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 (dr, dW = small, small sample where
force could be considered constant) To calculate work for a larger displacement,
you have to take an integral W = dW = F • dr
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
• 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)
We already know this technique manually with a graph; we can now call this ‘manual integration’
Tutorial on the Integral• Mathematically, the integral is used to calculate a sum
composed of many, many tiny parts.• In physics, the integral is used to find a measurable change
resulting from very small incremental changes. It’s useful to think of it as a fancy addition process or a multiplication process, depending on the situation.
• Here’s a specific example. Velocity times time gives displacement. If the velocity is changing with time, but a very tiny time change is used to calculate a very tiny displacement, we can nonetheless assume the velocity was constant during that tiny time change.
• If we calculate tiny displacements this way (recalculating our velocity for each time increment), then add the tiny displacements up to get a larger displacement, we have done “integration”. 21
Velocity as a Function of Time Displacement
• Velocity can be represented as follows:
• Rearrangement of this expression yields:
• What this means is that we can calculate a tiny displacement dx from the velocity v at a given time times a tiny time increment dt.
dx vdt
22
dxv
dt
Summing the Displacements
• When we sum up these tiny displacements, we use the following notation:
• This notation indicates that we are summing up all the little displacements dx starting at position xo at time to until we reach a final position and time, xf and tf. The velocity v may be a function of time, and may be slightly different for one time increment dt and the next time increment.
23
f f
o o
x t
x tdx vdt
Evaluating Integrals• We will evaluate polynomial integrals by reversing
the process we used in taking a polynomial derivative. This process is sometimes called “anti-differentiation”, or “doing an anti-derivative”.
• The general method for doing an anti-derivative is:
• Can you see how this is the reverse of taking a derivative? (Note: This “indefinite integral” requires us to add a constant C to compensate for constants that may have been lost during differentiation…but we will do “definite integrals” that do not require C.)
24
1
if then 1
nndx At
At x Cdt n
Evaluating Definite Integrals
• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)
25
2
2 2
4 1
23 2
41
2.03 2
1.0
Starting time: 1 s; ending time: 2 s.
Starting position:4 m
where 3 2 5
(3 2 5)
3 2 5
3 2 1
4 5
4 22 7 15
19
f
f
x
x
f
f
f
dxv v t t
dt
dx t t dt
t t tx
x t t t
x
x
Evaluating Definite Integrals
• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)
26
2
2 2
4 1
23 2
41
2.03 2
1.0
Starting time: 1 s; ending time: 2 s.
Starting position:4 m
where 3 2 5
(3 2 5)
3 2 5
3 2 1
4 5
4 22 7 15
19
f
f
x
x
f
f
f
dxv v t t
dt
dx t t dt
t t tx
x t t t
x
x
Evaluating Definite Integrals
• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)
27
2
2 2
4 1
23 2
41
2.03 2
1.0
Starting time: 1 s; ending time: 2 s.
Starting position:4 m
where 3 2 5
(3 2 5)
3 2 5
3 2 1
4 5
4 22 7 15
19
f
f
x
x
f
f
f
dxv v t t
dt
dx t t dt
t t tx
x t t t
x
x
Evaluating Definite Integrals
• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)
28
2
2 2
4 1
23 2
4
1
2.03 2
1.0
Starting time: 1 s; ending time: 2 s.
Starting position:4 m
where 3 2 5
(3 2 5)
3 2 5
3 2 1
4 5
4 22 7 15
19
f
f
x
x
f
f
f
dxv v t t
dt
dx t t dt
t t tx
x t t t
x
x
Evaluating Definite Integrals
• When a “definite integral” with “limits of integration” is to be evaluated, first do the “anti-derivative”. Then evaluate the resulting functions using the top limits, evaluate them again using the bottom limits, then subtract the two values to get the answer. (Note: do each side of the equation separately.)
29
2
2 2
4 1
23 2
4
1
23 2
1
Starting time: 1 s; ending time: 2 s.
Starting position:4 m
where 3 2 5
(3 2 5)
3 2 5
3 2 1
4 5
4 22 7 15
19
f
f
x
x
f
f
f
dxv v t t
dt
dx t t dt
t t tx
x t t t
x
x
• Sample problem: Consider a force that is a function of time:
F(t) = (3.0 t – 0.5 t2)N• If this force acts upon a 0.2 kg particle at rest for 3.0 seconds, what is the
resulting velocity and position of the particle?
30
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.
• Sample problem: Consider a force that is a function of time:
F(t) = (16 t2 – 8 t + 4)N• If this force acts upon a 4 kg particle at rest for 1.0 seconds, what is the
resulting change in velocity of the particle?
32
Try these samples on a clean sheet of paper.
• 1. Calculate the work done by a force that applies F(x) = 3x + x3, while it moves an object from 3 to 8 meters.
• 2. A tractor pulls a trailer with a force that varies according to F(x) = x – 3x. Calculate the work done from rest to 2.0 m.
• 3.
• Problem: Derive an expression for the work done by a spring as it is stretched from its equilibrium position
• 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?
Work Energy Theorem
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
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.)
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
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?
• 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?
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?
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
Units of Power
Watt = J/s ft lb / s horsepower
550 ft lb / s 746 Watts
• 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.
• 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%?
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.
• Problem: Using what you know about units, calculate how many Joules is in a kilowatt-hour.
Conservative and Non-Conservative Forces
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
Potential Energy, U 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
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
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.
•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
• 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.
Conservation of Mechanical Energy
System
Boundary
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.
Law of Conservation of Mechanical Energy
E = U + K= Constant
We only allow U and K to interchange.We ignore Eint (thermal energy)
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
h
Pendulum Energy
½mv12 + mgh1 = ½mv2
2 + mgh2 For any points two points in the pendulum’s swing
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
• 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?
•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?
Non-conservative Forces and Conservation of Energy
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
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.
• 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?
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
Stable Equilibrium – 1st and 2nd Derivatives
U
x
Unstable Equilibrium – 1st and 2nd Derivatives
U
x
Neutral Equilibrium – 1st and 2nd Derivatives
U
x
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
Molecular potential energy diagrams
R
U
• 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.
• 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).
Linear Momentum
October 31, 2008
Announcements
Turn in homework due today:Chapter 8, problems 28,29,31Next week, W-F, Rocket Project
Linear Momentum
Momentum is a measure of how hard it is to stop or turn a moving object.
p = mv (single particle) P = Σpi (system of particles)
• Problem: How fast must an electron move to have the same momentum as a proton moving at 300 m/s?
• Problem: A 90-kg tackle runs north at 5.0 m/s and a 75-kg quarterback runs east at 8.0 m/s. What is the momentum of the system of football players?
Impulse (J)
Impulse is the integral of force over a period of time.
J = Fdt The change in momentum of a particle is
equal to the impulse acting on it. p = J
Problem: Restate Newton’s 2nd Law in terms of impulse.
time
Force Impulsive forces are generally of high magnitude and short duration.
Problem: A 150-g baseball moving at 40 m/s 15o below the horizontal is struck by a bat. It leaves the bat at 55 m/s 35o above the horizontal. What is the impulse exerted by the bat on the ball?
If the collision took 2.3 ms, what was the average force?
• Problem: An 85-kg lumberjack stands at one end of a floating 400-kg log that is at rest relative to the shore of a lake. If the lumberjack jogs to the other end of the log at 2.5 m/s relative to the shore, what happens to the log while he is moving?
• Problem: Two blocks of mass 0.5 kg and 1.5 kg are placed on a horizontal, frictionless surface. A light spring is compressed between them. A cord initially holding the blocks together is burned; after this, the block of mass 1.5 kg moves to the right with a speed of 2.0 m/s. A) What is the speed and direction of the other block? B) What was the original elastic energy in the spring?
Conservation of Linear Momentum
The linear momentum of a system is conserved unless the system experiences an external force.
Collision Review
Collisions in 1 and 2 Dimension
Collision review
In all collisions, momentum is conserved. Elastic Collisions: No deformation occurs
Kinetic energy is also conserved.
Inelastic Collisions: Deformation occurs Kinetic energy is lost.
Perfectly Inelastic Collisions Objects stick together, kinetic energy is lost.
Explosions Reverse of perfectly inelastic collision, kinetic
energy is gained.
• 1 Dimensional Problem: A 1.5 kg cart traveling at 1.5 m/s collides with a stationary 0.5 kg cart and sticks to it. At what speed are the carts moving after the collision?
• 1 Dimensional Problem: A 1.5 kg cart traveling at 1.5 m/s collides elastically with a stationary 0.5 kg cart. At what speed are each of the carts moving after the collision?
• 1 Dimensional Problem: What is the recoil velocity of a 120-kg cannon that fires a 30-kg cannonball at 320 m/s?
For 2-dimensional collisions
Use conservation of momentum independently for x and y dimensions.
You must resolve your momentum vectors into x and y components when working the problem
•2 Dimensional Problem: A pool player hits a cue ball in the x-direction at 0.80 m/s. The cue ball knocks into the 8-ball, which moves at a speed of 0.30 m/s at an angle of 35o angle above the x-axis. Determine the angle of deflection of the cue ball.
Collision Problems Practice Day
Problem 1
A proton, moving with a velocity of vii, collides elastically with another proton initially at rest. If the two protons have equal speeds after the collision, find a) the speed of each in terms of vi and b) the direction of the velocity of each.
Center of Mass
Center of Mass
The center of mass of a system is the point at which all the mass can be assumed to reside.
Sometimes the system is an assortment of particles and sometimes it is a solid object.
Mathematically, you can think of the center of mass as a “weighted average”.
Center of Mass – system of points
Analogous equations exist for velocity and acceleration, for example
i icm
x mx
M
i icm
y my
M
i icm
z mz
M
,,
x i ix cm
v mv
M
Sample Problem Determine the Center of Mass.
2 4-2 0
0
2
-2
x
y
2 kg
3 kg1 kg
Center of Mass-- solid objects
If the object is of uniform density, you pick the geometric center of the object.
xx
x For combination objects,
pick the center of each part and then treat each center as a point.
x
x
Center of Mass Problem Determine the
center of mass of this shape.
x
y
2R
Center of Mass for more complicated situations If the shapes are points or simple geometric
shapes of constant density, then
cm
xdmx
M
i icm
x mx
M
Anything else is more complicated, and
• Problem: Find the x-coordinate of the center of mass of a rod of length L whose mass per unit length varies according to the expression = x.
Lx
y
• Problem: A thin strip of material of mass M is bent into a semicircle of radius R. Find its center of mass.
x
y
R
R
-R
Motion of a System of Particles
Motion of a System of Particles
If you have a problem involving a system of many particles, you can often simplify your problem greatly by just considering the motion of the center of mass.
If all forces in the sytem are internal, the motion of the center of mass will not change.
The effect of an external force that acts on all particles can be simplified by considering that it acts on the center of mass; that is:
Fext = Macm
Homework problem – collaborative work (15 minutes) This problem is a toughie! However, it gets easier if you
use the idea that the center of mass of the system of Romeo, Juliet, and canoe does not change because all forces are internal. The question is paraphrased below:
Romeo (77 kg) sits in the back of a canoe 2.70 m away from Juliet (55 kg) who sits in the front. The canoe has a mass of 80 kg. Juliet moves to the rear of the canoe to kiss Romeo. How far does the canoe move forward when she does this? (Assume a canoe that is symmetrical).