Physics 111: Mechanics Lecture 6

Wenda Cao

NJIT Physics Department

October 7-13, 2013

Chapter 6 Work and Kinetic Energy

6.1 Work 6.2 Kinetic Energy and the Work-Energy

Theorem 6.3 Work and Energy with Varying Forces 6.4 Power

October 7-13, 2013

Why Energy? Why do we need a concept of energy? The energy approach to describing

motion is particularly useful when Newton’s Laws are difficult or impossible to use.

Energy is a scalar quantity. It does not have a direction associated with it.

October 7-13, 2013

Kinetic Energy Kinetic Energy is energy associated with

the state of motion of an object For an object moving with a speed of v

SI unit: joule (J) 1 joule = 1 J = 1 kg m2/s2

2

2

1mvK

October 7-13, 2013

Kinetic Energy for Various Objects

2

2

1mvKE

October 7-13, 2013

Why ?2

2

1mvK

October 7-13, 2013

Work W

Start with Work “W”

Work provides a link between force and energy Work done on an object is transferred to/from it If W > 0, energy added: “transferred to the

object” If W < 0, energy taken away: “transferred from

the object”

xFmvmv x 20

2

2

1

2

1

October 7-13, 2013

Definition of Work W The work, W, done by a constant force on an

object is defined as the product of the component of the force along the direction of displacement and the magnitude of the displacement

F is the magnitude of the force Δ x is the magnitude of the object’s displacement is the angle between

xFxFW

)cos(

and F x

October 7-13, 2013

Work Unit This gives no information about

the time it took for the displacement to occur the velocity or acceleration of the object

Work is a scalar quantity SI Unit

Newton • meter = Joule N • m = J J = kg • m2 / s2 = ( kg • m / s2 ) • m

xFxFW

)cos(

xFmvmv )cos(2

1

2

1 20

2

October 7-13, 2013

Work: + or -? Work can be positive, negative, or zero.

The sign of the work depends on the direction of the force relative to the displacement

Work positive: if 90°> > 0° Work negative: if 180°> > 90° Work zero: W = 0 if = 90° Work maximum if = 0° Work minimum if = 180°

sFsFW

)cos(

October 7-13, 2013

Example: When Work is Zero

A man carries a bucket of water horizontally at constant velocity.

The force does no work on the bucket

Displacement is horizontal Force is vertical cos 90° = 0

xFW )cos(

October 7-13, 2013

Example: Work Can Be Positive or Negative

Work is positive when lifting the box

Work would be negative if lowering the box The force would still be

upward, but the displacement would be downward

October 7-13, 2013

Work Done by a Constant Force

The work W done a system by an agent exerting a constant force on the system is the product of the magnitude F of the force, the magnitude Δr of the displacement of the point of application of the force, and cosθ, where θ is the angle between the force and displacement vectors: cosrFrFW

F

II

F

IIIr

F

I

r

F

IVr

r

0IW

cosrFWIV rFWIII

rFWII

October 7-13, 2013

Work and Force An Eskimo returning pulls a sled as shown.

The total mass of the sled is 50.0 kg, and he exerts a force of 1.20 × 102 N on the sled by pulling on the rope. How much work does he do on the sled if θ = 30° and he pulls the sled 5.0 m ?

J

mN

xFW

2

2

102.5

)0.5)(30)(cos1020.1(

)cos(

October 7-13, 2013

Work Done by Multiple Forces

If more than one force acts on an object, then the total work is equal to the algebraic sum of the work done by the individual forces

Remember work is a scalar, so this is the algebraic sum

net by individual forcesW W

rFWWWW FNgnet )cos(

October 7-13, 2013

Kinetic Energy Kinetic energy associated with the

motion of an object Scalar quantity with the same unit as

work Work is related to kinetic energy

2

2

1mvK

xFmvmv net 20

2

2

1

2

1

KKKW ifnet

October 7-13, 2013

October 7-13, 2013

Work-Energy Theorem When work is done by a net force on an

object and the only change in the object is its speed, the work done is equal to the change in the object’s kinetic energy

Speed will increase if work is positive Speed will decrease if work is negative

20

2

2

1

2

1mvmvWnet

KKKW ifnet

October 7-13, 2013

Problem Solving Strategy Identify the initial and final positions of the body,

and draw a free body diagram showing and labeling all the forces acting on the body

Choose a coordinate system List the unknown and known quantities, and

decide which unknowns are your target variables Calculate the work done by each force. Be sure to

check signs. Add the amounts of work done by each force to find the net (total) work Wnet

Check whether your answer makes sense

October 7-13, 2013

Work and Kinetic Energy The driver of a 1.00103 kg car traveling on the

interstate at 35.0 m/s slam on his brakes to avoid hitting a second vehicle in front of him, which had come to rest because of congestion ahead. After the breaks are applied, a constant friction force of 8.00103 N acts on the car. Ignore air resistance. (a) At what minimum distance should the brakes be applied to avoid a collision with the other vehicle? (b) If the distance between the vehicles is initially only 30.0 m, at what speed would the collisions occur?

October 7-13, 2013

Work and Kinetic Energy (a) We know Find the minimum necessary stopping distance

Nfkgmvsmv k33

0 1000.8,1000.1,0,/0.35

233 )/0.35)(1000.1(2

1)1000.8( smkgxN

22

2

1

2

1iffricNgfricnet mvmvWWWWW

202

10 mvxfk

mx 6.76

October 7-13, 2013

Work and Kinetic Energy (b) We know Find the speed at impact. Write down the work-energy theorem:

Nfkgmvsmv k33

0 1000.8,1000.1,0,/0.35

Nfkgmsmvmx k33

0 1000.8,1000.1,/0.35,0.30

xfm

vv kf 22

02

22

2

1

2

1ifkfricnet mvmvxfWW

2233

22 /745)30)(1000.8)(1000.1

2()/35( smmN

kgsmv f

smv f /3.27

October 7-13, 2013

Work with Varying Forces On a graph of force as a function

of position, the total work done by the force is represented by the area under the curve between the initial and the final position

Straight-line motion

Motion along a curve

...... bbxaax xFxFW

dxFWx

x x 2

1

ldFdlFdlFWP

P

P

P

P

P

2

1

2

1

2

1

cos ||

October 7-13, 2013

Work-Energy with Varying Forces

Work-energy theorem Wtol = K holds for varying forces as well as for constant ones

dx

dvv

dt

dx

dx

dv

dt

dva x

xxx

x

dxdx

dvmvdxmadxFW

x

x

xx

x

x x

x

x xtot 2

1

2

1

2

1

KmvmvWtot 21

22 2

1

2

1

x

v

v xtot dvmvW 2

1

October 7-13, 2013

Spring Force: a Varying Force

Involves the spring constant, k

Hooke’s Law gives the force

F is in the opposite direction of x, always back towards the equilibrium point.

k depends on how the spring was formed, the material it is made from, thickness of the wire, etc. Unit: N/m.

dkF

October 7-13, 2013

Measuring Spring Constant Start with spring at its

natural equilibrium length. Hang a mass on spring

and let it hang to distance d (stationary)

From

so can get spring constant.

April 19, 2023

0xF kx mg

mgk

d

October 7-13, 2013

Work done on a Spring To stretch a spring, we must do

work We apply equal and opposite

forces to the ends of the spring and gradually increase the forces

The work we must do to stretch the spring from x1 to x2

Work done on a spring is not equal to work done by a spring

21

22 2

1

2

12

1

2

1

kxkxdxkxdxFWx

x

x

x x

October 7-13, 2013

Power Work does not depend on time interval The rate at which energy is transferred is

important in the design and use of practical device

The time rate of energy transfer is called power

The average power is given by

when the method of energy transfer is work

WP

t

October 7-13, 2013

Instantaneous Power Power is the time rate of energy transfer.

Power is valid for any means of energy transfer

Other expression

A more general definition of instantaneous power

vFt

xF

t

WP

vFdt

rdF

dt

dW

t

WP

t

0lim

cosFvvFP

October 7-13, 2013

Units of PowerThe SI unit of power is called the

watt 1 watt = 1 joule / second = 1 kg . m2 / s3

A unit of power in the US Customary system is horsepower 1 hp = 550 ft . lb/s = 746 W

Units of power can also be used to express units of work or energy 1 kWh = (1000 W)(3600 s) = 3.6 x106 J

October 7-13, 2013

A 1000-kg elevator carries a maximum load of 800 kg. A constant frictional force of 4000 N retards its motion upward. What minimum power must the motor deliver to lift the fully loaded elevator at a constant speed of 3 m/s?

Power Delivered by an Elevator Motor

yynet maF ,

0 MgfT

NMgfT 41016.2

W

smNFvP4

4

1048.6

)/3)(1016.2(

hpkWP 9.868.64