Work and Energy

1. Work Energy Work done by a constant force

(scalar product)

Work done by a varying force

(scalar product & integrals)

2. Kinetic Energy

Chapter 6: Work and 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 ?

W [Joule] = ( F cos ) d

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

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

= 37.0o) by a person. A friction force Ff =50.0 N is exerted to the crate. Determinethe work done by each force acting on thecrate.

Work and Energy

Example 1A (cont’d)

WP = FP d cos ( 37o )

Wf = Ff d cos ( 180o )

Wg = m g d cos ( 90o )

WN = FN d cos ( 90o )

180o

90o

d

F.B.D.

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 = Wi

= 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 - F x (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 - F x (? m) = - m (33.3 m/s)2 / 2

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

Work and Energy

Forces

Forces on a hammerhead

Work and Energy

Work and Energy

Work and Energy

Spring Force (Hooke’s Law)

FS(x) = - k x

FPFS

Natural Length x > 0

x < 0

Spring Force(Restoring Force):The spring exerts its force in thedirection opposite the displacement.

Work and Energy

Work Done to Stretch a Spring

x2

W = FP(x) dxx1

FS(x) = - k x

W

Natural Length

FPFS

Work and Energy

Work and Energy

lb

W = F|| dl la

Work Done bya Varying Force

l 0

Work and Energy

Example 1A

A person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the person do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the person do ?

Work and Energy

(a) Find the spring constant k 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 xmax

2 = 1.1 J

(c) x2 = 0.030 mWP = FP(x) d x = 1.1 J x1 = 0

Example 1A (cont’d)

Work and Energy

Example 1B

A person pulls on the spring, stretching it3.0 cm, which requires a maximum force of 75 N. How much work does the spring do ? If, instead, theperson compressesthe spring 3.0 cm,how much workdoes the spring do ?

Work and Energy

(a) Find the spring constant k 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 mWS = FS(x) d x = -1.1 J x1 = 0

Example 1B (cont’d)

Work and Energy

Example 2

A 1.50-kg block is pushed against a spring(k = 250 N/m), compressing it 0.200 m, andreleased. What will be the speed of theblock when it separates from the spring at

x = 0? Assume k =0.300.

(i) F.B.D. first !(ii) x < 0

FS = - k x

Work and Energy

(a) The work done by the spring is

(b) Wf = - kFN (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) m v2 – 0 v = 2.34 m/s

x2 = 0 mWS = FS(x) d x = +5.00 J x1 = -0.200 m

Example 2 (cont’d)

Work and Energy