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Work and Energy. Unit 4. W = Fcos q D x. Lesson 1 : Work Done by a Constant Force. When a force acts on an object while displacement occurs, the force has done work on the object.

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Work and Energy

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Work and energy

Work and Energy

Unit 4


Work and energy

W = Fcosq Dx

Lesson 1 : Work Done by a Constant Force

When a force acts on an object while displacement occurs, the force has done work on the object.

The magnitude of work (W) is the product of the amount of the force applied along the direction of displacement and the magnitude of the displacement.


Work and energy

N . m = Joule (J)

+ Work

- Work

If the force has a component in the direction of the displacement.

If the force has a component in the opposite direction of the displacement.

Units of Work

Determining the Sign of Work


Work and energy

Rank the following situations in order of the work done by the force on the object, from most positive to most negative. [Displacement is to the right and of the same magnitude.]

Example 1


Work and energy

4.0 kg

5.0 m

m = 0.30

30o

Find the work done by all forces as a 4.0 kg mass slides 5.0 m down a 30o incline where the coefficient of kinetic friction is 0.30.

Example 2


Work and energy

F (N)

x (m)

Work done is the area under the graph

Graphical Analysis of Work

F

WF = FDx

Dx


Work and energy

Since

W = Fcosq Dx

and

A . B = ABcosq

then

W = F .Dx

Work is a Scalar (Dot) Product


Work and energy

A particle moving in the xy plane undergoes a displacement Dx = (2.0 i + 3.0 j) m as a constant force F = (5.0 i + 2.0 j) N acts on the particle.

^

^

^

^

Example 3

a) Calculate the magnitudes of the displacement and the force.

b) Calculate the work done by force F.


Work and energy

xf

W = SFxDx

xi

Lesson 2 : Work Done by a Varying Force


Work and energy

xf

ò

lim S FxDx =

Fxdx

Dx  0

xi

xf

ò

W =

Fxdx

xi

As Dx approaches 0,

Therefore,


Work and energy

A force acting on a particle varies with x, as shown above. Calculate the work done by the force as the particle moves from x = 0 to x = 6.0 m.

Example 1


Work and energy

The interplanetary probe shown above is attracted to the Sun by a force given by

1.3 x 1022

F = -

x2

Example 2


Work and energy

This equation is in SI units, where x is the Sun-probe separation distance. Determine how much work is done by the Sun on the probe as the probe-Sun separation changes from 1.5 x 1011 m to 2.3 x 1011 m.


Work and energy

~ 60 squares

Graphical Solution

Each square = (0.05 N)(0.1 x 1011 m) = 5 x 108 J


Work and energy

Hooke’s Law

force exerted by spring

Fs = -kx

position relative to equilibrium position

spring constant in N/m

Work Done by a Spring

Negative sign signifies that the force exerted by spring is always directed opposite to the displacement.


Work and energy

stretched spring

equilibrium position

compressed spring


Work and energy

xf

ò

(-kx)dx =

Ws = ½ kx2

ò

Ws =

Fsdx =

xi

½ kx2

Work done by the spring force is positive because the force is in the same direction as displacement.


Work and energy

xf

ò

(-kx)dx = ½ kxi2 - ½ kxf2

Ws =

xi

xf

xf

ò

ò

Ws =

Fappdx =

kxdx =

½ kxf2 - ½ kxi2

xi

xi

Generalized Work Done by Spring

Generalized Work Done on Spring


Work and energy

Example 3

A common technique used to measure the spring constant (k) is shown above. The spring is hung vertically, and an object of mass m is attached to its lower end. Under the action of the “load” mg, the spring stretches a distance d from its equilibrium position.


Work and energy

a) If a spring is stretched 2.0 cm by a suspended mass of 0.55 kg, what is the spring constant of the spring ?

b) How much work is done by the spring as it stretches through this distance ?

c) Suppose the measurement is made on an elevator with an upward vertical acceleration a. Will the unaware experimenter arrive at the same value of the spring constant ?


Work and energy

Example 4

If it takes 4.00 J of work to stretch a Hooke’s Law spring 10.0 cm from its unstressed length, determine the extra work required to stretch it an additional 10.0 cm.


Work and energy

Example 5

A light spring with spring constant 1200 N/m is hung from an elevated support. From its lower end a second light spring is hung, which has spring constant 1800 N/m. An object of mass 1.50 kg is hung at rest from the lower end of the second spring.

a) Find the total extension distance of the pair of springs.


Work and energy

b) Find the effective spring constant of the pair of springs as a system. We describe these springs as in series.


Work and energy

Example 6

^

^

A force F = (4xi + 3yj) N acts on an object as the object moves in the x-direction from the origin to x = 5.00 m.

Find the work W = F . dx done on the object by the force.

ò


Work and energy

Work done by SF is

xf

ò

SW =

SF dx

xi

Lesson 3 : Work-Kinetic Energy Theorem


Work and energy

(by chain-rule)

dx

dv

dv

vf

xf

xf

xf

ò

ò

ò

ò

dt

dx

dt

SW =

SW =

SW =

SW =

m dx

ma dx

mv dv

m dx

SW =

½ mvf2 – ½ mvi2

vi

xi

xi

xi


Work and energy

KE = ½ mv2

SW = KEf – KEi = DKE

Kinetic Energy

Work - Kinetic Energy Theorem

If work done on a system only changes its speed, the work done by the net force equals the change in KE of the system.


Work and energy

Example 1

A 6.0 kg block initially at rest is pulled to the right along a horizontal, frictionless surface by a constant horizontal force of 12 N. Find the speed of the block after it has moved 3.0 m.


Work and energy

Example 2

A man wishes to load a refrigerator onto a truck using a ramp. He claims that less work would be required to load the truck if the length L of the ramp were increased. Is his statement valid ?


Work and energy

Example 3

A 4.00 kg particle is subject to a total force that varies with position as shown above. The particle starts from rest at x = 0. What is its speed at

a) x = 5.00 m

b) x = 10.00 m

c) x = 15.00 m


Work and energy

vf - vi

ax =

t

Lesson 4 : Situations Involving Kinetic Friction

SFx = max

(SFx)Dx = (max)Dx

Dx = ½ (vi + vf) t


Work and energy

(

vf - vi

)

(SFx)Dx = m

½ (vi + vf) t

t

This is not work because Dx is displacement of a particle – the book is not a particle !

-fkDx = DKE

(SFx)Dx = ½ mvf2 – ½ mvi2

(SFx)Dx = -fkDx = ½ mvf2 – ½ mvi2 = DKE


Work and energy

DKE = -fkd + SWother forces

d = length of any path followed

-fkd = DKE

KEf = KEi - fkd + SWother forces

DKE in General

OR


Work and energy

Example 1

A 6.0 kg block initially at rest is pulled to the right along a horizontal surface by a constant horizontal force of 12 N.

a) Find the speed of the block after it has moved 3.0 m if the surfaces in contact have a coefficient of kinetic friction of 0.15.


Work and energy

b) Suppose the force F is applied at and angle q as shown below. At what angle should the force be applied to achieve the largest possible speed after the block has moved 3.0 m to the right ?


Work and energy

DEint = fkd

Change in Internal Energy due to Friction

The result of a friction force is to transform KE into internal energy, and the increase in internal energy is equal is equal to the decrease in KE.

DEsystem = DKE + DEint = 0

-fkd + DEint = 0


Work and energy

Example 2

A 40.0 kg box initially at rest is pushed 5.00 m along a rough, horizontal floor with a constant applied horizontal force of 130 N. If the coefficient of friction between box and floor is 0.300, find

a) the work done by the applied force

b) the increase in internal energy in the box-floor system due to friction


Work and energy

c) the work done by the normal force

d) the work done by the gravitational force


Work and energy

e) the change in kinetic energy of the box

f) the final speed of the box.


Work and energy

Lesson 5 : Power

Same amount of work done

Time interval is different


Work and energy

W

P =

Dt

W

dW

P = lim

=

Dt  0

Dt

dt

dW

F . dx

F . v

P =

=

=

dt

dt

Average Power

time rate of energy transfer

Instantaneous Power


Work and energy

1 W = 1 J/s = 1 kg . m2/s3

1 hp = 746 W

Units of Power

SI unit of power is J/s or the Watt (W).

In the U.S. customary system, the unit of power is the horsepower (hp).


Work and energy

1 kWh = (103 W)(3600 s) = 3.60 x 106 J

The kilowatt-hour (kWh)

The energy transferred in 1 h at the constant rate of 1kW = 1000 J/s.

* Note that a kWh is a unit of energy, not power.


Work and energy

An elevator car has a mass of 1600 kg and is carrying passengers having a combined mass of 200 kg. A constant friction force of 4000 N retards its motion upward, as shown above.

Example 1


Work and energy

a) What power delivered by the motor is required to lift the elevator car at a constant speed of 3.00 m/s ?

b) What power must the motor deliver at the instant the speed of the elevator is v if the motor is designed to provide the elevator car with an upward acceleration of 1.00 m/s2 ?


Work and energy

Example 2

Find the instantaneous power delivered by gravity to a 4 kg mass 2 s after it has fallen from rest.


Work and energy

Example 3

Find the instantaneous power delivered by the net force at t = 2 s to a 0.5 kg mass moving in one dimension according to x(t) = 1/3 t3.


Work and energy

Consider a car of mass m that is accelerating up a hill, as shown above. An automotive engineer measures the magnitude of the total resistive force to be ft = (218 + 0.70v2) N where v is in m/s. Determine the power the engine must deliver to the wheels as a function of speed.

Example 4


Work and energy

100 kg

The 100 kg box shown above is being pulled along the x-axis by a student. The box slides across a rough surface, and its position x varies with time t according to the equation x = 0.5t3 + 2t, where x is in meters and t is in seconds.

Example 5 : AP 2003 #1

a) Determine the speed of the box at time t = 0.


Work and energy

b) Determine the following as functions of time t.

i. The kinetic energy of the box.

ii. The net force acting on the box.

iii. The power being delivered to the box.


Work and energy

d) Indicate below whether the work done on the box by the student in the interval t = 0 to t = 2 s would be greater than, less than, or equal to the answer in part c). Justify your answer.

____Greater than ____ Less than ____ Equal to

c) Calculate the net work done on the box in the interval t = 0 to t = 2 s.


Work and energy

system = book + Earth

DKE = 0

(vi = 0, vf = 0)

Lesson 6 : Potential Energy

Work done on system by external agent in lifting book

When book is at yb, the energy of the system has potential to become KE.


Work and energy

When lifting at constant velocity,

^

^

W = (Fapp) .Dx = (mgj) . [(yb – ya)j] = mgyb - mgya

Ug = mgy

W = DUg

Gravitational Potential Energy

Units for Ug are Joules (J). Like work and KE, Ug is a scalar quantity.


Work and energy

Example 1

A bowling ball held by a careless bowler slips from the bowler’s hands and drops on the bowler’s toe. Choosing floor level as the y = 0 point of your coordinate system, estimate the change in gravitational PE of the ball-Earth system as the ball falls. Repeat the calculation, using the top of the bowler’s head as the origin of coordinates.


Work and energy

Example 2

A 400 N child is in a swing that is attached to ropes 2.00 m long. Find the gravitational potential energy of the child-Earth system relative to the child’s lowest position when

a) the ropes are horizontal

b) the ropes make a 30o angle with the vertical

c) the child is at the bottom of the circular arc.


Work and energy

As book falls from yb to ya, the work done by the gravitational force on the book is

^

^

Won book = (mg) . (Dx) = (-mg j) . [(ya – yb) j]

Lesson 7 : Conservation of Mechanical Energy


Work and energy

From the work-kinetic energy theorem,

Won book = DKEbook

So,

DKE = -DUg

Won book = mgyb - mgya

DKEbook = mgyb - mgya

For the book-Earth system,

mgyb – mgya = -(mgya – mgyb) = -(Uf – Ui) = -DUg


Work and energy

Bringing DU to left side of the equation,

This sum of KE and Ug is called mechanical energy.

Emech = KE + U

represents all types of potential energy

Conservation of Mechanical Energy (isolated, frictionless system)

KEf + Uf = KEi + Ui

DKE + DUg = 0

(KEf – KEi) + (Uf – Ui) = 0


Work and energy

Us = ½ kx2

Elastic potential energy stored in a spring

Elastic Potential Energy

WFapp = ½ kxf2 – ½ kxi2


Work and energy

A ball of mass m is dropped from a height h above the ground, as shown.

Example 1

a) Neglecting air resistance, determine the speed of the ball when it is at a height y above the ground.


Work and energy

b) Determine the speed of the ball at y if at the instant of release it already has an initial upward speed vi at the initial altitude h.


Work and energy

A pendulum consists of a sphere of mass m attached to a light cord of length L. The sphere is released from rest at point A when the cord makes an angle qA with the vertical, and the pivot at P is frictionless.

Example 2


Work and energy

a) Find the speed of the sphere when it is at the lowest point B.

b) What is the tension TB in the cord at B ?


Work and energy

The launching mechanism of a toy gun consists of a spring of unknown spring constant. When the spring is compressed 0.120 m, the gun, when fired vertically, is able to launch a 35.0 g projectile to a maximum height of 20.0 m above the position of the projectile before firing.

Example 3

a) Neglecting all resistive forces, determine the spring constant.


Work and energy

b) Find the speed of the projectile as it moves through the equilibrium position of the spring (where xB = 0.120 m).


Work and energy

A bead slides without friction around a loop-the-loop. The bead is released from a height h = 3.50R.

Example 4

a) What is its speed at point A ?


Work and energy

b) How large is the normal force on it if its mass is 5.00 g ?


Work and energy

An object of mass m starts from rest and slides a distance d down a frictionless incline of angle q. While sliding, it contacts an unstressed spring of negligible mass as shown above. The object slides an additional distance x as it is brought momentarily to rest by compression of the spring (of spring constant k). Find the initial separation d between object and spring.

Example 5


Work and energy

Example 6 : AP 1989 # 1

A 0.1 kg block is released from rest at point A as shown above, a vertical distance h above the ground. It slides down an inclined track, around a circular loop of radius 0.5 m, then up another incline that forms an angle of 30o with the horizontal. The block slides off the track with a speed of 4 m/s at point C, which is a height of 0.5 m above the ground. Assume the entire track to be frictionless and air resistance to be negligible.


Work and energy

b) On the figure below, draw and label all the forces acting on the block when it is at point B, which is 0.5 m above the ground.

a) Determine the height h.


Work and energy

c) Determine the magnitude of the force exerted by the track on the block when it is at point B.

d) Determine the maximum height above the ground attained by the block after it leaves the track.


Work and energy

e) Another track that has the same configuration, but is NOT frictionless, is used. With this track it is found that if the block is to reach point C with a speed of 4 m/s, the height h must be 2 m. Determine the work done by the frictional force.


Work and energy

Example 7 : AP 1985 # 2

An apparatus to determine coefficients of friction is shown above. The box is slowly rotated counter-clockwise. When the box makes an angle q with the horizontal, the block of mass m just starts to slide, and at this instant the box is stopped from rotating. Thus at angle q, the block slides a distance d, hits the spring of force constant k, and compresses the spring a distance x before coming to rest.


Work and energy

In terms of the given quantities, derive an expression for each of the following.

a) ms, the coefficient of static friction


Work and energy

b) DE, the loss in total mechanical energy of the block-spring system from the start of the block down the incline to the moment at which it comes to rest on the compressed spring


Work and energy

c) mk, the coefficient of kinetic friction


Work and energy

Lesson 8 : Conservative and Nonconservative Forces

Conservative Forces

1. The work done by a conservative force on a particle moving between any two points is independent of the path taken by the particle.

2. The work done by a conservative force on a particle moving through any closed path is zero. (A closed path is one in which the beginning and end points are identical.)


Work and energy

yi

Fg

Wg depends on y coordinates and is independent of the path

yf

Wg is zero when the object moves over any closed path (where yi = yf).

Fg

Examples of Conservative Forces

a) Gravitational Force

Wg = mgyi - mgyf


Work and energy

Ws depends on y coordinates and is independent of the path

Wg is zero when the object moves over any closed path (where yi = yf).

b) Force exerted by a spring

Ws = ½ kxi2 – ½ kxf2


Work and energy

Nonconservative forces acting within a system cause a change in the mechanical energy of the system.

Nonconservative Forces

A force that does not satisfy the properties of a conservative force.

Work done by force depends on the path.


Work and energy

Friction force is a nonconservative force.

If book is displaced along blue path, work done against friction is less than if book is pushed along curved brown path.


Work and energy

DEmech = DKE + DU = -fkd

If the forces acting on objects within a system are conservative, then the mechanical energy of the system is conserved.

If some of the forces acting on objects within a system are nonconservative, then the mechanical energy of the system changes.

If a friction force acts within a system,


Work and energy

Example 1

A 3.00 kg crate slides down a ramp. The ramp is 1.00 m in length and inclined at an angle of 30.0o. The crate starts from rest at the top, experiences a constant friction force of magnitude 5.00 N, and continues to move a short distance on the horizontal floor after it leaves the ramp. Use energy methods to determine the speed of the crate at the bottom of the ramp.


Work and energy

Diagram for Example 1


Work and energy

A child of mass m rides on an irregularly curved slide of height h = 2.00 m. The child starts from rest at the top.

Example 2


Work and energy

a) Determine his speed at the bottom, assuming no friction is present.

b) If a force of kinetic friction acts on the child, how much mechanical energy does the system lose ? Assume that vf = 3.00 m/s and m = 20.0 kg.


Work and energy

Two blocks are connected by a light string that passes over a frictionless pulley. The block of mass m1 lies on a horizontal surface and is connected to a spring of force constant k. The system is released from rest when the spring is unstretched. If the hanging block of mass m2 falls a distance h before coming to rest, calculate the coefficient of kinetic friction between the block m1 and the surface.

Example 3


Work and energy

xf

ò

Wc =

Fx dx = -DU

xi

xf

ò

DU = Uf – Ui = -

Fx dx

xi

Lesson 9 : Conservative Forces and PE

The work done by a conservative force equals the decrease in PE of the system.

DU = is negative when Fx and dx are in the same direction.


Work and energy

F

equiibrium

DUg is negative

DUs is negative

xf

ò

Uf (x) = -

Fx dx + Ui

xi

F


Work and energy

dU

Fx = -

dx

The x-component of a conservative force acting on an object within a system equals the negative derivative of the potential energy of the system with respect to x.

dU = -Fx dx


Work and energy

dUs

dUg

Fs = -

Fg = -

dx

dy

d

d

Fg = -

Fs = -

(1/2 kx2)

(mgy)

dy

dx

Fs = -kx

(Hooke’s Law)

Gravitational PE

Elastic PE

Fg = -mg


Work and energy

Example 1

Consider the potential energy of two molecules given by

A

B

U =

-

r12

r6

Find the force along the line joining the two molecules.


Work and energy

KE

Negative slope equals F

PE

“Stable” equilibrium

U(x) is a minimum

Lesson 10 : Energy Diagrams


Work and energy

dU

F =

= -kx

dx

The force at a given point is the negative slope of the curve.

Where the graph reaches maxima or minima, the force will be 0.

Stable equilibrium points will be located at the minima.


Work and energy

“unstable” equilibrium

Fx is negative

Fx is positive

Acceleration away from x = 0

Acceleration away from x = 0


Work and energy

Example 1

For the potential energy curve shown below,


Work and energy

a) determine whether the force Fx is positive, negative, or zero at the five points indicated.

b) indicate points of stable, unstable, and neutral equilibrium.


Work and energy

c) sketch the curve for Fx vs. x from x = 0 to x = 9.5 m


Work and energy

Example 2

A particle moves along a line where the potential energy of its system depends on its position r as graphed below. In the limit as r increases without bound, U(r) approaches +1J.


Work and energy

a) Identify each equilibrium position for this particle. Indicate whether each is a point of stable, unstable, or neutral equilibrium.

b) The particle will be bound if the total energy of the system is in what range ?


Work and energy

Now suppose that the system has energy -3J. Determine

c) the range of positions where the particle can be found.

d) its maximum kinetic energy.


Work and energy

e) the location where it has maximum kinetic energy.

f) the binding energy of the system – that is, the additional energy that it would have to be given in order for the particle to move out to r  infinity .


Work and energy

Example 3

Jane, whose mass is 50.0 kg, needs to swing across a river (having width D) filled with man-eating crocodiles to save Tarzan from danger. She must swing into a wind exerting constant horizontal force F, on a vine having length L and initially making an angle q with the vertical.


Work and energy

Taking D = 50.0 m, F = 110 N, L = 40.0 m, and q = 50.0o,

a) with what minimum speed must Jane begin her swing in order to just make it to the other side ?


Work and energy

b) Once the rescue is complete, Tarzan and Jane must swing back across the river. With what minimum speed must they begin their swing ? Assume that Tarzan has a mass of 80.0 kg.


Work and energy

Example 4 : AP 1987 # 2

The following graph shows the potential energy U(x) of a particle as a function of its position x.

a) Identify all points of equilibrium for this particle.


Work and energy

b) Determine the kinetic energy of the particle at the following positions :

i. x = 2.0 m

Suppose the particle has a constant total energy of 4.0 J, as shown by the dashed line on the graph.

ii. x = 4.0 m


Work and energy

c) Can the particle reach the position x = 0.5 m ? Explain.

d) Can the particle reach the position x = 5.0 m ? Explain.


Work and energy

e) On the grid below, carefully draw a graph of the conservative force acting on the particle as a function of x, for 0<x<7 m.


Work and energy

Example 5 : AP 1995 # 2

A particle of mass m moves in a conservative force field described by the potential energy function U(r) = a(r/b + b/r), where a and b are positive constants and r is the distance from the origin. The graph of U(r) has the following shape.


Work and energy

a) In terms of the constants a and b, determine the following :

i. The position ro at which the potential energy is a minimum.

ii. The minimum potential energy Uo.


Work and energy

b) Sketch the net force on the particle as a function of r on the graph below, considering a force directed away from the origin to be positive, and a force directed toward the origin to be negative.


Work and energy

The particle is released from rest at r = ro/2.

c) In terms of Uo and m, determine the speed of the particle when it is at r = ro.

d) Write the equation or equations that could be used to determine where, if ever, the particle will again come to rest. It is not necessary to solve for this position.


Work and energy

e) Briefly and qualitatively describe the motion of the particle over a long period of time.


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