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Chapter 7: Work and EnergyPowerPoint Presentation

Chapter 7: Work and Energy

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Chapter 7: Work and Energy

- Work Energy
Work done by a constant force

(scalar product)

Work done by a varying force

(scalar product & integrals)

- Kinetic Energy

Work-Energy Theorem

Work and Energy

Forms of Mechanical Energy

Work and Energy

Work and Energy

Work by a Baseball Pitcher

A baseball pitcher is doing work on

the ball as he exerts the force over

a displacement.

v1 = 0

v2 = 44 m/s

Work and Energy

Work Done by a Constant Force (I)

Work(W)

How effective is the force in moving a body ?

Both magnitude (F) and directions (q ) must be taken into account.

W[Joule] = ( F cos q ) d

Work and Energy

Work Done bya Constant Force (II)

Example:Work done on the bag by the person..

Special case: W = 0 J

a) WP = FP d cos ( 90o )

b) Wg = m g d cos ( 90o )

Nothing to do with the motion

Work and Energy

Example 1A

A 50.0-kg crate is pulled 40.0 m by a

constant force exerted (FP = 100 N and

q = 37.0o) by a person. A friction force Ff =

50.0 N is exerted to the crate. Determine

the work done by each force acting on the

crate.

Work and Energy

Example 1A (cont’d)

F.B.D.

WP = FP d cos ( 37o )

Wf = Ff d cos ( 180o)

Wg = m g d cos ( 90o)

WN = FN d cos ( 90o)

180o

d

90o

Work and Energy

Work-Energy Theorem

Wnet= Fnet d = ( m a ) d

= m [ (v2 2 – v1 2 ) / 2d ] d

= (1/2) m v2 2 – (1/2) m v1 2

= K2 – K1

Work and Energy

Example 2

A car traveling 60.0 km/h to can brake to

a stop within a distance of 20.0 m. If the car

is going twice as fast, 120 km/h, what is its

stopping distance ?

(a)

(b)

Work and Energy

Example 2 (cont’d)

(1)Wnet = F d(a) cos 180o

= - F d(a) = 0 – m v(a)2 / 2

- Fx (20.0 m) = -m (16.7 m/s)2 / 2

(2) Wnet = F d(b) cos 180o

= - F d(b) = 0 – m v(b)2 / 2

- Fx (? m) = -m (33.3 m/s)2 / 2

(3)F & m are common. Thus, ? = 80.0 m

Work and Energy

Spring Force (Hooke’s Law)

FS

Spring Force

(Restoring Force):

The spring exerts its force in the

direction opposite

the displacement.

FP

x > 0

Natural Length

x < 0

FS(x) = - k x

Work and Energy

Work Done to Stretch a Spring

FS

FP

FS(x) = - k x

Natural Length

x2

W = FP(x) dx

x1

W

Work and Energy

Example 1A

A person pulls on the spring, stretching it

3.0 cm, which requires a maximum force

of 75 N. How much work does the person

do ? If, instead, the

person compresses

the spring 3.0 cm,

how much work

does the person do ?

Work and Energy

Example 1A (cont’d)

- (a) Find the spring constantk
- k = Fmax / xmax
- = (75 N) / (0.030 m) = 2.5 x 103 N/m

- (b) Then, the work done by the person is
- WP = (1/2) k xmax2 = 1.1 J
- (c)

x2 = 0.030 m

WP = FP(x) d x = 1.1 J

x1 = 0

Work and Energy

Example 1B

A person pulls on the spring, stretching it

3.0 cm, which requires a maximum force

of 75 N. How much work does the spring

do ? If, instead, the

person compresses

the spring 3.0 cm,

how much work

does the spring do ?

Work and Energy

Example 1B (cont’d)

- (a) Find the spring constantk
- k = Fmax / xmax
- = (75 N) / (0.030 m) = 2.5 x 103 N/m

- (b) Then, the work done by the spring is
- (c) x2 = -0.030 m WS = -1.1 J

x2 = -0.030 m

WS = FS(x) d x = -1.1 J

x1 = 0

Work and Energy

Example 2

A 1.50-kg block is pushed against a spring

(k = 250 N/m), compressing it 0.200 m, and

released. What will be the speed of the

block when it separates from the spring at

x = 0? Assume mk =

0.300.

FS = - k x

(i) F.B.D. first !

(ii) x < 0

Work and Energy

Example 2 (cont’d)

(a) The work done by the spring is

(b) Wf = - mkFN (x2 – x1) = -4.41 (0 + 0.200)

(c) Wnet = WS+ Wf = 5.00 - 4.41 x 0.200

(d) Work-Energy Theorem: Wnet=K2 – K1

4.12 = (1/2) mv2 – 0

v = 2.34 m/s

x2 = 0 m

WS = FS(x) d x = +5.00 J

x1 = -0.200 m

Work and Energy

Potential Energy and Energy Conservation

- Conservative/Nonconservative Forces
Work along a path

(Path integral)

Work around any closed path

(Path integral)

- Potential Energy

Mechanical Energy Conservation

Energy Conservation

Work Done bythe Gravitational Force (I)

Near the Earth’s surface

l

(Path integral)

Energy Conservation

Work Done bythe Gravitational Force (II)

Near the Earth’s surface

(Path integral)

dl

Energy Conservation

Wg < 0 if y2 > y1

Wg > 0 if y2 < y1

The work done by the gravitational

force depends only on the initial and

final positions..

Work Done bythe Gravitational Force (III)Energy Conservation

Wg(ABCA)

=Wg(AB) +

Wg(BC) +

Wg(CA)

=mg(y1 – y2) +

0 +

mg(y2- y1)

= 0

Work Done bythe Gravitational Force (IV)C

B

dl

A

Energy Conservation

Wg = 0 for a closed path

The gravitational force is a conservative force.

Work Done bythe Gravitational Force (V)Energy Conservation

Work Done by Ff (I)

(Path integral)

- μmg L

L depends on the path.

LB

Path B

Path A

LA

Energy Conservation

The work done by the friction force

depends on the path length.

The friction force:

(a) is a non-conservative force;

(b) decreases mechanical energy of the system.

Work Done by Ff (II)Wf = 0 (any closed path)

Energy Conservation

Example 1

A 1000-kg roller-coaster car moves from

point A, to point B and then to point C.

What is its gravitational potential energy

at B and C

relative to

point A?

Energy Conservation

Wg(ABC) =Wg(AB) + Wg(BC)

=mg(yA- yB) + mg(yB - yC)

= mg(yA - yC)

Wg(AC) = Ug(yA) – Ug(yC)y

B

A

dl

B

C

A

Energy Conservation

Climbing the Sear tower

Work and Energy

Work and Energy

The Burj Khalifa is the largest

man made structure in the world

and was designed by Adrian

Smith class of 1966

thebatt.com Febuary 25th

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