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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

Unit 4


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.


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


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


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


F (N)

x (m)

Work done is the area under the graph

Graphical Analysis of Work

F

WF = FDx

Dx


Since

W = Fcosq Dx

and

A . B = ABcosq

then

W = F .Dx

Work is a Scalar (Dot) Product


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.


xf

W = SFxDx

xi

Lesson 2 : Work Done by a Varying Force


xf

ò

lim S FxDx =

Fxdx

Dx  0

xi

xf

ò

W =

Fxdx

xi

As Dx approaches 0,

Therefore,


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


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

1.3 x 1022

F = -

x2

Example 2


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.


~ 60 squares

Graphical Solution

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


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.


stretched spring

equilibrium position

compressed spring


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.


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


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.


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 ?


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.


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.


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


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 done by SF is

xf

ò

SW =

SF dx

xi

Lesson 3 : Work-Kinetic Energy Theorem


(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


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.


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.


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 ?


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


vf - vi

ax =

t

Lesson 4 : Situations Involving Kinetic Friction

SFx = max

(SFx)Dx = (max)Dx

Dx = ½ (vi + vf) t


(

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


DKE = -fkd + SWother forces

d = length of any path followed

-fkd = DKE

KEf = KEi - fkd + SWother forces

DKE in General

OR


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.


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 ?


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


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


c) the work done by the normal force

d) the work done by the gravitational force


e) the change in kinetic energy of the box

f) the final speed of the box.


Lesson 5 : Power

Same amount of work done

Time interval is different


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


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).


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.


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


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 ?


Example 2

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


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.


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


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.


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.


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.


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.


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.


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.


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.


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


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


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


Us = ½ kx2

Elastic potential energy stored in a spring

Elastic Potential Energy

WFapp = ½ kxf2 – ½ kxi2


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.


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.


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


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 ?


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.


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


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 ?


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


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


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.


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.


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.


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.


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.


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

a) ms, the coefficient of static friction


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


c) mk, the coefficient of kinetic friction


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.)


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


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


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.


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.


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,


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.


Diagram for Example 1


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


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.


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


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.


F

equiibrium

DUg is negative

DUs is negative

xf

ò

Uf (x) = -

Fx dx + Ui

xi

F


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


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


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.


KE

Negative slope equals F

PE

“Stable” equilibrium

U(x) is a minimum

Lesson 10 : Energy Diagrams


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.


“unstable” equilibrium

Fx is negative

Fx is positive

Acceleration away from x = 0

Acceleration away from x = 0


Example 1

For the potential energy curve shown below,


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.


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


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.


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 ?


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.


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 .


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.


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 ?


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.


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.


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


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

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


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.


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.


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.


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.


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.


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


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