Chapter 5 Work and Energy

Definition of Work

There is a difference between the ordinary definition of work and the scientific definition of work

Ordinary Definition: To do something that takes physical or mental effort

Scientific definition: Work is equal to the magnitude of the applied force times the displacement of an object

What is necessary for work to be done? A force that causes displacement of an object

does work on the object Therefore, work is not done on an object

unless the object is moved because of the action of a force

Work is being done!!!

What is necessary for work to be done? Work is done only when components of a

force are parallel to a displacement When the force on an object and the object’s

displacement are perpendicular, no work is done

Force

Displacement

No Work was Done!!!

What about forces at an angle?

Only the component of force that is parallel to the direction of the object’s displacement does work. Example: A person pushes a box across a

frictionless floor

FBD For the Box

Fapp

Fapp,x

Fapp,y

FN

Fg

What part of the applied force is parallel to the displacement?

Displacement of Box

General Equation For Work Done

Work= Component of Force that does the work x displacement x cos (angle between the force vector and the displacement)

cosFdW

Net Work done by a Constant Force

If there are many constant forces acting on the object, you can find the net work done by finding the net force acting on the object

Net work= Net force x displacement x cos of the angle between them

cosdFW netnet

Angles between vectors

Vector Orientation Angle between them

Cos(θ)

Θ= 90 Cos(90) =0

Θ= 0 Cos(0) = 1

Θ= 180 Cos(180) = -1

Units for Work

The unit for Work is the Joule

I J= 1 Nm One Joule = One Newton x One meter

The sign of work is important

Work is a scalar quantity, but it can be positive or negative

Work is negative when the force is in the direction opposite the displacement For example, the work done by the frictional

force is always negative because the frictional force is opposite the displacement

Sample Problem p. 193 # 10

A flight attendant pulls her 70 N flight bag a distance of 253 m along a level airport floor at a constant velocity. The force she exerts is 40.0 N at an angle of 52.0° above the horizontal. Find the following: The work she does on the flight bag The work done by the force of friction on the

flight bag The coefficient of kinetic friction between the

flight bag and the floor

FBD

Fg

Fapp,y

Fapp,x

Ff

FappFN

Work done by flight attendant

Only the component of Fapp that is parallel to the displacement does work. Fapp,x is parallel to the displacement Fapp,x = 40cos52 = 24.63 N

Remember that in the W=Fdcosθ equation, θ represents the angle between the force vector and the displacement vector

W= (24.63 N)(253 m) cos(0)= 6230 J

Fapp,x d Θ = 0°

Work done by friction

The bag is moving at constant velocity, so what is Ff? Ff = Fapp,x= 40cos(52)= 24.62 N

W= Fdcosθ= (24.62N)(253m)(cos180)

= -6230 J

Ffd Θ =180°

Find μk

What is FN?

Fn + Fapp,y = Fg

Fn = Fg - Fapp,y= 70N- 40sin(52)= 38.48 N

N

fk F

F

64.48.38

63.24

N

fk F

F

Energy- Section 5.2 p. 172

Kinetic Energy- The energy of an object due to its motion

I don’t Have Kinetic Energy

I Have KineticEnergy

Kinetic Energy Depends on Speed and Mass

Kinetic energy = ½ x mass x speed2

The unit for KE is Joules (J)

2

2

1mvKE

Sample Problem p. 173

A 7.00 kg bowling ball moves at 3.0 m/s. How much kinetic energy does the bowling ball have? How fast must a 2.45 g tennis ball move in order to have the same kinetic energy as the bowling ball?

Solve the Problem

KE= ½ mv^2= ½ (7kg)(3m/s)^2= 31.5 J

How fast must 2.45 g ball move to have the same KE? Convert g to kg 2.45 g = .00245 kg Solve for v

s

m

kg

J

m

KEv 160

00245.

)5.31(22

Work-Kinetic Energy Theorem

The net work done by a net force acting on an object is equal to the change in kinetic energy of the object

You must include all the forces acting on the object for this to work!

ifnet KEKEKEW

Sample problem p. 176 #2

A 2000 kg car accelerates from rest under the actions of two forces. One is a forward force of 1140 N provided by the traction between the wheels and the road. The other is a 950 N resistive force due to various frictional forces. Use the work-KE theorem to determine how far the car must travel for its speed to reach 2.0 m/s.

What information do we have?

M= 2000 kg Vi= 0m/s Vf= 2 m/s

Fforward= 1140 NFfriction= 950 N

What is the Work-Ke Theorem?

Remember that:

So Expand the equation to this:

ifnet KEKEKEW

22

2

1

2

1cos ifnet mvmvdF

cosdFW netnet 22

2

1

2

1if mvmvKE

Solve for Fnet Fnet= Fforward- Ffriction= 190 N forward

Rearrange the equation for d and plug in values

Why is θ= 0 in the denominator? Because the net force is in the same direction as the displacement.

mF

mvmvd

net

if

210cos190

)0)(2000(21

2200021

cos21

21 2222

22

2

1

2

1cos ifnet mvmvdF

Potential Energy Section 5.2

Potential Energy is stored energy

There are two types of PE Gravitational PE Elastic PE

Gravitational Potential Energy

Gravitational Potential Energy (PEg) is the energy associated with an object’s position relative to the Earth or some other gravitational source

heightx s

m9.81 x mass

2mghPEg

Elastic Potential Energy

Elastic Potential Energy (PEelastic) is the potential energy in a stretched or compressed elastic object.

k= spring (force) constant X= displacement of spring

2

2

1kxPEelastic

Displacement of Spring

Sample Problem p. 180 #2

The staples inside a stapler are kept in place by a spring with a relaxed length of 0.115 m. If the spring constant is 51.0 N/m, how much elastic potential energy is stored in the spring when its length is 0.150 m?

What do we know?

K = 51.0 N/m

We need to get x, in order to use the equation for elastic PE

X is the distance the spring is stretched or compressed

Relaxed length is 0.115 m, stretched length is 0.150. How much was it stretched? 0.150- 0.115 m= 0.035 m= x

Solve the problem

JkxPEelastic 031.035.0.512

1

2

1 22

Conservation of Energy – 5.3

The total amount of energy in the universe is a constant

So we say that energy is conserved

From IPC: The Law of Conservation of Energy: Energy can neither be created nor destroyed

Mechanical Energy

There are many types of energy (KE, PE, Thermal, etc)

We are concerned with Mechanical Energy

Mechanical Energy is the sum of kinetic energy and all forms of potential energy

ME= KE + PE

Conservation of ME

In the absence of friction, mechanical energy is conserved

When friction is present, ME can be converted to other forms of energy (i.e. thermal energy) so it is not conserved.

fi MEME

Expanded Form of Conservation of ME Without elastic PE

With elastic PE

ffii mghmvmghmv 22

2

1

2

1

2222

2

1

2

1

2

1

2

1fffiii kxmghmvkxmghmv

Practice Problem p. 185 #2

A 755 N diver drops from a board 10.0 m above the water’s surface. Find the diver’s speed 5.00 m above the water’s surface. Find the diver’s speed just before striking the water.

What do we know?

W= 755 N Initial height = 10 m Vi= 0 m/s

There is no elastic PE involved.

Solve part a.

What is the diver’s speed 5.0 m above the water’s surface?

M= Weight/g=76.96 kg Vi= 0m/s Initial height = 10 m Final Height = 5 m

Rearrange equation and solve for vf

fif mghmghmv 2

2

1

s

m

m

mghmghv fif 10

77

)5)(81.9)(77()10)(81.9)(77((22

ffii mghmvmghmv 22

2

1

2

1Vi = 0 m/s

Second Part

What is the diver’s speed just before striking the water?

M= Weight/g=76.96 kg Vi= 0m/s Initial height = 10 m Final Height = 0 m

Finish the Problem

ffii mghmvmghmv 22

2

1

2

1Vi = 0 m/s

2

2

1fi mvmgh

hf = 0

s

mgh

m

mghv i

if 14)10)(81.9(22

2

Power- Section 5.4

Power:The rate at which work is done

The unit for power is Watts

1 Watt = 1 J

1 s

Time

Work

t

WP

Alternate Form for Power

Speed x ForceFvP

Sample Problem (Not in book)

At what rate is a 60 kg boy using energy when he runs up a flight of stairs 10 m high in 8.0 s?

Time = 8 s

What is work done? W=Fdcos(θ)

Time

Work

t

WP

Solve the Problem

What force does the boy apply to get himself up the stairs? F= Weight= mg= 588.6 N d= 10m

W= Fdcos(θ)=588.6N(10m)(cos(0)) W=5886 J

P=W/t = 5886 J/ 8s= 735.8 Watts