PH 201-4A spring 2007PH 201 4A spring 2007

Work and EnergyWork and Energy

Lectures 16-17

Chapter 6(Cutnell & Johnson, Physics 7th edition)

1

Work and Energy: Work done by a constant force

• Constant pushing force F pointing in the same direction as the resulting displacement S• It takes work to push a stalled car.• Force and displacement are two essential elements of work. W increases when F pincreases and S increases• Work is defined as the magnitude of the force F times the magnitude of the displacement W = F x S• Work does not convey directional information, it is a scalar quantityy y• Unit of measurement for work

Force x Distance = WorkNewton (N) x meter (m) = joule (J)

• important feature: The idea of work is tied up with the idea of motion. If there is no movement of the object, the work done by the force acting on the object is zero

2

The force is directed at angle θ relative to the displacementThe work done on an object by a constant force F is W = (Fcosθ)s where F is the magnitude of the force s is the magnitude of the displacement θ is the angle betweenmagnitude of the force, s is the magnitude of the displacement, θ is the angle between the force and displacement.

The force component that points along the displacement is Fcosθ and it is used in defining work.

The force component perpendicular to the displacement does no work. To do work, there must be a force and a displacement and since there is no displacement in the perpendicular direction there is no work done by the perpendicular component of the forceforce.

3

Positive and Negative Work. Bench-pressing

During the lifting phase, the force F does positive work

W = (Fcosθ)s θ = 0° cosθ = 1

If the force component points in the direction opposite to the displacement, the work is negative.

During the lowering phaseDuring the lowering phase

W = (Fcosθ)s = (Fcos180°)s== -Fs

Since cos180° = -1

4

Accelerating a CrateA 120kg crate on the flatbed of a truck is moving with an acceleration a = +1.5m/s2 along the positive x axis The crate does not slip with respect to the truck as the truckthe positive x axis. The crate does not slip with respect to the truck, as the truck undergoes a displacement s = 65m. What is the total work done on the crate by all the forces acting on it?

1.2x104 N

5

Work done by a variable force

The curved plot of Fcosθ vs S has been approximated by straight line Plot of Fcosθ vs S for a force segments. This is a good approximation if ΔS→0.

W=(Fcosθ)1⋅ΔS1+…+(Fcosθ)n ⋅ΔSn=

that varies with position. The work done by this force as a particle moves displacement S equals the area under the plot

sum of rectangular areas=

area under the graph

equals the area under the plot.

6

WorkProblem 11: A 1200kg car is being driven up a 5.0° hill. The frictional force is directed opposite to the motion of the car and has a magnitude of fk = 5.0x102N. The force F is pp g kapplied to the car by the road and propels the car forward. In addition to those two forces, two other forces act on the car: its weight W, and the normal force N directed perpendicular to the road surface. The length of the hill is 3.0x102m. What should be the magnitude of F so that the net work done by all the forces acting on the car is g y g+150,000 J?

7

The Work-Energy Theorem and Kinetic Energy• When a net force performs work on an object, there is always a result from the effort –change in the kinetic energy.change in the kinetic energy.• The relationship that relates work to the change in kinetic energy is the Work-Energy Theorem.• A constant net external force ΣF acts on an airplane of mass m• For simplicity it has the same direction as displacement so s p c ty t as t e sa e d ect o as d sp ace e t s

Newton’s 2nd Law: net force produces acceleration a = ΣF/m; (ΣF)s = mas• Speed of the plane changes from v0 to vf

• from kinematics: vf2 = v0

2 + 2as => as = ½(vf2 – v0

2) => (ΣF)s = 1/2mvf2 – 1/2mv0

2

• The kinetic energy KE of an object with mass m and speed v is defined as KE = 1/2mv2

• The work-energy theorem: when a net external force does work W on an object, the kinetic energy of the object changes from initial value KE0 to the final value KEf, the difference between the two values being equal to the work

W = KEf – KE0 = 1/2mvf2 – 1/2mv0

2

8

The Work-Energy Theorem and Kinetic EnergyProblem 12: A 0.075kg arrow is fired horizontally. The bowstring exerts an average force of 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave theof 65 N on the arrow over a distance of 0.90 m. With what speed does the arrow leave the bow?

9

The work-energy theorem deals with the work done by the net external force. The work-energy theorem does not apply to the work done by an individual force. If W>0 then KE increases; if W<0 then KE decreases; if W=0 then KE remains constant.

Downhill Skiing: A 58kg skier is coasting down a 25° slope. A kinetic friction force fk=70N opposes her motion. Near the top of the slope, the skier’s speed is v0=3.6m/s. Ignoring air resistance, determine the speed vf at a point that is displaced 57m downhill.

10

Work and Kinetic Energy

In a circular orbit the gravitational force F is always perpendicular to the displacement s of the satellite and does no work

KE = constant

W>0KE increases

In an elliptical orbit, there can be a component of the force along the displacement

Work is done

W<0KE decreases 11

The Work-Energy Theorem and Kinetic EnergyThe speed of a hockey puck decreases from 45.00 to 44.67m/s in coasting 16m across the ice. Find the coefficient of kinetic friction between the puck and the ice.p

12

Work done by a force of gravity

• basketball of mass m moving vertically downwardvertically downward• mg is the only force acting• initial height h0, final hf• the displacement s is downward

s=h -hs=h0-hf• Wgravity = (Fcosθ)s = mg(h0-hf)

Valid for any path taken between the initial and final height.

Since only the difference between h0 and hf appears in the equation, the vertical distances themselves need not be measured from the earth. For example, they would be measured relative to a zero level that is one meter above the ground. h0-hf will be be easu ed e at e to a e o e e t at s o e ete abo e t e g ou d 0 f bethe same.

13

Gravitational Potential Energy• An object in motion has kinetic energy• An object may possess energy by virtue of its position relative to the earth – such an j y p gy y pobject is said to have gravitational potential energy

Wgravity = mgh0 – mghf

• The gravitational potential energy PE is the energy that an object of mass m has by

Initial gravit. potential energy PE0

Final gravit. potential energy PEf

• The gravitational potential energy PE is the energy that an object of mass m has by virtue of its position relative to the surface of the earth. That position is measured by the height h of the object relative to an arbitrary zero level.

PE = mgh SI unit: joule (J)PE = mgh SI unit: joule (J)

• The gravitational potential energy belongs to the object and the earth as a system although one often speaks of the object alone as possessing the gravitational potential energyenergy.

14

Gravitational Potential Energy

ymgmghmghWUPE if Δ=−=−=Δ≡Δ (1)ymgmghmghWUPE ifgg ΔΔ≡Δ

The choice of the reference point for the potential energy is completely arbitrary.It is customary to locate the reference point where the gravitational force is zero.

It is possible to show that for this reference point

(1)

It is possible to show that for this reference point

Earthhi

hf −=

dr

mGMU Eg

Δy

RE

ME

Ug⎟⎟⎠

⎞⎜⎜⎝

⎛ −=⎟

⎟⎠

⎞⎜⎜⎝

⎛−−=Δ

frr

rrmGM

rrmGMU

and

i

ifE

ifEg

11m(2)

y

0

RE rg ⎠⎝⎠⎝ fiif

Where r is the distance between the center of the Earth and the object.

L t h th t f ti l f di l d th h

E

E

RmGM− Let us show that for a particle of mass m displaced through a

small vertical distance Δy near the Earth’s surface the general expression (2) for the change in gravitational energy reduces to a familiar relationship (1). If both the initial and final positions of the p ( ) pparticle are close to the Earth’s surface, then rf-ri=Δy and rf⋅ri=RE

2

ymgyR

mGMUE

Eg Δ=Δ≈Δ 2

15

Gravitational Potential Energy

Sign arises from the choice of the zero reference point (where U=0)zero reference point (where U=0) which is r = ∞

Gravitational potential energy of a particle of mass 1kg gravitationally attracted by another particle of mass 1kg.g

Total mechanical energy:E = K + U = 1/2mv2 – GMm/r = constantIf the only force is gravity.y g y

For a circular orbit of planets:v2 = GMs/r; K = 1/2mv2 = GMsm/2rE = 1/2mv2 – GMsm/r = - GMsm/2rs sGMsm/2r + (-GMsm/r) = - GMsm/2r

16

The motor of a Scout rocket uses up all the fuel and stops when the rocket is at an altitude of 200km above the surface of the Earth and is moving vertically at 8.50km/s. How high will this rocket rise? Ignore any residual atmospheric friction.

17

Example: A meteoroid (a chunk of rock) is initially at rest in interplanetary space at a large distance from the Sun. Under the influence of gravity, the meteoroid begins to fall toward the Sun along a straight radial line. With what speed does it strike the Sun?

18

Conservative Forces, Nonconservative Forces and the Work-Energy Theorem

An important property of the gravitational force: when an object is moved from one• An important property of the gravitational force: when an object is moved from one place to another, the work done by the gravitational force does not depend on the choice of path. For this reason it is called a conservative force.• Definition:Version 1: A force is conservative when the work it does on a moving object isVersion 1: A force is conservative when the work it does on a moving object is independent on the path between the object’s initial and final positions.Version 2: A force is conservative when it does no net work on an object moving around a closed path starting and finishing at the same point.

• A force is nonconservative if the work it does on an object moving between two points depends on the path of the motion between the points.

Kinetic friction force example of nonconservative forceKinetic friction force – example of nonconservative force

• In normal conditions, conservative and nonconservative forces (friction, air resistance) act simultaneously on an object• Work done by the net force:• Work done by the net force:

W = Wc + Wnc

19

Wc + Wnc = 1/2mvf2 – 1/2mv0

2The Work-Energy Theorem

Wc = Wgravity = mg(h0 – hf)

mg(h0 – hf) + Wnc = 1/2mvf2 – 1/2mv0

2

Wnc = (1/2mvf2 – 1/2mv0

2) + (mghf – mgh0)

Wnc = (KEf – KE0) + (PEf – PE0)Wnc (KEf KE0) (PEf PE0)Net work done by nonconserv. forces

Change to kinetic energy

Change in potential energy

Wnc = ΔKE + ΔPE

The conservation of mechanical energy:

forces

gy

E = KE + PE - total mechanical energy

20

Conservation of Mechanical Energy• E = KE + PE total mechanical energy• W = (KE KE ) + (PE PE ) = (KE + PE ) (KE + PE )• Wnc = (KEf – KE0) + (PEf – PE0) = (KEf + PEf) – (KE0 + PE0)The net work done by external nonconservative forces changes the total mechanical energy from an initial value E0 to a final value of Ef• Suppose Wnc = 0 => Ef = E0 1/2mvf

2 + mghf = 1/2mv0 + mgh0total mechanical energy conserved

Yes

total mechanical energy conserved

Yes

No

The total mechanical energy (E = KE + PE) of an object remains constant as the object moves provided that the net work done by external nonconservative forces is zeromoves provided that the net work done by external nonconservative forces is zero, Wnc = 0 J

21

Conservation of Mechanical EnergyE = KE + PE = constant if Wnc = 0

If friction and wind resistance are ignored, a bobsled run illustrates how kinetic and t ti l b i t t d hil th t t l h i l ipotential energy can be interconverted while the total mechanical energy remains

constant. The total mechanical energy is all potential energy at the top and all kinetic energy at the bottom.

22

A golf ball of mass 50g released from a height of 1.5m above a concrete floor bounces back to a height of 1.0m

a) What is the kinetic energy of the ball just before contact with the floor begins? Ignore i f i tiair friction.

b) What is the kinetic energy of the ball just after contact with the floor ends?c) What is the loss of energy during contact with the floor?

23

A pendulum consists of a mass “m” tied to one end of a string of length “l”. The other end of the string is attached to a fixed point on the wall. Suppose that the pendulum is initially held at an angle of 90° with the vertical. If the pendulum is released from this position what will be the speed of the pendulum at the instant it passes through itsposition, what will be the speed of the pendulum at the instant it passes through its lowest point?

24

In a rollercoaster, a car starts on the top of a 30m high mountain. It rolls down into a valley and then up a 20m high mountain. What is the speed of the car at the bottom of the valley, at ground level? What is the speed of the car at the top of the second mountain?

25

A block of mass m slides down an inclined plane into a loop of radius R. a) Neglecting friction, what is the minimum speed the block must have at the highest point of the loop to stay in the loop. b) At what vertical height on the inclined plane (in terms of radius of the loop) must the block be released if it is to have the required minimum speed at the top of the loop?

26

Nonconservative Forces and the Work-Energy TheoremProblem 17: A 55.0 kg skateboarder starts out with a speed of 1.80 m/s. He does +80.0 J

of work on himself by pushing his foot against the ground. In addition, friction does -265 J of work on him. In both cases, the forces doing the work are nonconservative. The final speed of the skateboarder is 6.00 m/s.

a) ΔPE = PEf – PE0?b) How much has the vertical height of the skater changed and is the skater above or

below the starting point?

27

Nonconservative Forces and the Work-Energy TheoremProblem 54: A 3.00 kg model rocket is launched vertically straight up with sufficient initial speed to reach a maximum height of 1.00x102 m even though air resistance (aspeed to reach a maximum height of 1.00x10 m even though air resistance (a nonconservative force) performs -8.00X102 J of work on the rocket. How high would the rocket have gone without air resistance?

28

Power

• Car is moving up the hill• The important characteristic of the engine is not• The important characteristic of the engine is not how much force it can exert, but how much work it can perform in a given amount of time.

• The force is of secondary importance, because by shifting to a low gear we can make y p y g gsure that even a “weak” engine exerts enough force on the wheels to propel us uphill.• Rate of work is crucial – how fast the engine can propel the car up the hill.• When we move uphill, the gravitational force takes energy from the car (performs negative work). To keep the car moving the engine must perform an equal amount of g ) g gpositive work.• If the engine performs work at a fast rate, it can propel the car at a fast speed.• The rate at which a force does work on a body is called power delivered by the force.• If the force does the amount of work W in time Δt, average power: P = W/Δt

instantaneous power: P = W/Δt, Δt -> 0SI unit of power: 1 watt = 1 W = 1 J/sengineering units: 1 horsepower = 1 hp = 746 W1 kilowatt-hour = 1 KWh = 1000 W x 3600 s = 3.6x106J

Kilowatt-hour is used to measure electric energy delivered to homes and factories

29

Example:An elevator cage has a mass of 1000kg. How many horsepower must the motor deliver to the elevator if it is to raise the elevator cage at the rate of 2.0m/s?

30

A car accelerates uniformly from rest to 27 m/s in 7.0s along a level stretch of road. Ignoring friction, determine the average power required to accelerate the car ifa) The weight of the car is 1.2x104N, and b) weight is 1.6x104N

31

Other forms of energy• If the forces acting on a particle are conservative – mechanical energy is conserved• If nonconservative forces do work on the particle – mechanical energy is not conserved. This energy never disappears. It is converted into other forms of energy.• Chemical and nuclear energy: Chemical – KE and PE of electrons within atomNuclear – KE and PE of protons and neutrons within nuclei of atoms.• Thermal energy – KE and PE of atoms of the body.• Electric and magnetic energy – are forms of energy associated with electric charges and with light and radio waves• All these forms of energy can be transformed into one another. In any such transformative process, the sum of all the energies of all the pieces of matter involved in the process remains constant.

General law of conservation of energy: The form of energy changes, but the total amount of energy does not change.

32