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Chapter 6: 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 by a Baseball Pitcher. A baseball pitcher is doing work on

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Chapter 6: Work and Energy

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Chapter 6: 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

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

Example 1A (cont’d)

WP = 3195 [J]

Wf = -2000 [J] (< 0)

Wg = 0 [J]

WN = 0 [J]

180o

Work and Energy

Example 1A (cont’d)

Wnet = SWi

= 1195[J] (> 0)

• The body’s speed increases.

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

Forces on a hammerhead

Forces

Work and Energy

Work and Energy

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

FS

FP

FS(x) = - k x

Natural Length

x2

W = FP(x) dx

x1

W

Work and Energy

Work and Energy

lb

W = F||dl

la

Dl  0

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

Work and Energy