Chapter 5Work and Energy

Work and Energy

A force that causes a displacement of an object does work on the object.

W = Fd*Force in the direction of displacement

Perpendicular forces do not do work.

If an object is moved by a force other than horizontal…..

θ

F

d

Only the horizontal component causes displacement.

W = Fdcosθ Units: N * m = joules (J)

* Sample 5A pg 169

•Figure 5-3 Positive and Negative Work

Pg 170

Kinetic Energy -> Energy of Motion - depends on speed & mass

½ mv2 = KE

Pg. 172 derivation of Wnet = F Δ x = maΔx Wnet = ½ mvf

2 – ½ mvi2

x = d

Sample 5B pg 173

Work – Kinetic Energy TheoremW net = Δ KE

W net = KEf – KEi

W net = ½ mvf2 – ½ mvi

2

Net Work = Change in Kinetic Energy

Sample 5C pg 175

Potential Energy -> stored energy- potential to move due to position

Gravitational P.E. ->depends on height from a zero level

Unit: Joule = JP.E. g = mgh

Elastic P.E. -> depends on distance compressed or stretched

P.E. elastic = ½ kx2k = spring constantx = distance compressed or stretched

Figure 5-B pg 178Sample 5D pg 179

Conservation of Energy

Energy Remains constant !

Mechanical Energy -> is the sum of K.E. & all forms of P.E. associated with an object or group of objects (a system)

ME = KE + PE = ½ mv 2 + mgh

Energy

Non-MechanicalMechanical

Kinetic Potential

Gravitational Elastic

Conservation of Mechanical Energy

In a closed or isolated system the sum of the P.E. & K.E. remains constant; energy can change from P.E. K.E., but the sum remains the same .

(P.E. + K.E.)i = (P.E. + K.E.)f

Conservation of Energy Equation:

½ mvi2 + mghi = ½ mvf

2 + mghf

MASS ON TABLEmh1

h2

0

m

v2

m

v3 h3 = 0

PE1 + KE1 = PE2 + KE2 = PE3 + KE3

mgh1 + ½ m(0)2 = mgh2+ ½ mv22 = mg(0) + ½ mv3

2

mgh1 +0 = mgh2 + ½ mv22 = 0 + ½ mv3

2

v1 = 0

MASS ON TABLEmh1

h2

0

m

v2

m

v3 h = 0

PE1 + KE1 = PE2 + KE2 = PE3 + KE3

mgh1 + ½ m(0)2 = mgh2+ ½ mv22 = mg(0) + 1/2mv3

2

mgh1 +0 = mgh2 + ½ mv22 = 0 + ½ mv3

2

mgh1 = mgh2 + ½ mv22 = ½ mv3

2

v1 = 0

MASS ON TABLEmh1

h2

0

m

v2

m

v3 h = 0

PE1 + KE1 = PE2 + KE2 = PE3 + KE3

mgh1 + ½ m(0)2 = mgh2+ ½ mv22 = mg(0) + 1/2mv3

2

mgh1 +0 = mgh2 + ½ mv22 = 0 + ½ mv3

2

mgh1 = mgh2 + ½ mv22 = ½ mv3

2

gh1 = gh2 + ½ v22 = ½ v3

2

v1 = 0

* You would utilize any 2 equations depending on the situation

Sample Problem Pg. 184

MEi = MEf

PEi + KEi = PEf + Kef

mgh1 + ½ mv12 = mgh2 + ½ mv2

2

736 J + 0 J = 0 J + (.5)(25kg)vf2

736J/12.5kg=vf2

vf = 7.67 m/s* ME not conserved in the Presence of Friction

Fig. 5-12 pg 186

Rate of Work POWER!!!!!

P = W/Δt = (Fd)/Δt = F(d/t) = Fv = ma(d/t)

Units for Power: Watt = J / sec or hp = 746 W

* Sample Problem 5F pg. 188

Hmwk #1 Book Chp. 55A pg. 170 1,3,4 (d=1.1m)5B pg. 174 1,3,55C pg. 176 1,55D pg.180 1,2 (PE=.031 J)5E pg. 185 1,55F pg. 189 1,3

Hmwk #2 Workbook Chp. 55A 1. d= 195m 5D 1. m= 159kg

4. d= 20m 3. m= 83.8 kg 6. F= 400N 5. h= 2.09m 10. W= 20364 J 7. PE= 1.6 x 105 J

5B 1. m= 66kg 5E 3. hf =2970m 3. m= 67kg 5. 300 km/h (? m/s) 6. V= 1.084 m/s KE= 32J 8. KE= 82300 J 24:1

5C 2. KEf= 1760 J 5F 1. W= 2 x1011 J 4. 81m/s ? Km/hr 3. t= 48.5 s 6. m= 70 kg 5. P= 300W

Chapter Six

Momentum

Momentum p

Vector quantityProduct of an object’s mass & velocity p = mv

Units: kg * m/s**Sample 6A pg 209A change in Momentum takes force & time

F = maF = m (v/t)F = (mv)/tF = Δp/t

FΔt

Impulse – Momentum Theorem

Units: N*sec

Impulse

ExamplesTime >Force * Punch in a fight Time< Force * Karate Chop

Ft or ft = same p (change in momentum)

Sample 6B pg 211

*How does stopping time Relate to p = FΔt?

Sample 6C pg 212

Hmwk #3 BK & WKBK Chp. 6 Momentum A-C

Book Pg. 209 6A 1, 2 #1. 2482 kg m/s right #2 a. 120 kg m/s NW b. 94 kg m/s NW c. 27 kg m/s NW

Pg. 211 6B 1,3 #1 380N is to the left #3 -16 N sec (to the south)

Pg. 213 6C 1,2 #2 a. 14m/s N

b. 42m N c. 8 sec

WKBK

6-A: 1. 59 kg 3. 83.2 m/s 6. 4.17 x 10-2 kg m/s

6-B: 1. 1.8 sec 4. 11 m/s 8. 219 N up

6-C: 1. Δp= - 6 x 107 kg m/s Δx= 32m 3. Δp= 2.44 x 104 kg m/s Δx= 88.6 m east 7. 54 km

Einstein's General Relativity: compares measurements between two frames

of reference moving relative to each other.

Special Relativity: examines the measurements at relative speeds near the speed of light. E=mc2

ie. Objects moving near the speed of light will be shorter in the direction of motion, be more massive and have slower clocks than measurements made by the moving object

RELATIVE MOTION to different Frames of Reference:Motion with respect to high speed motionTime dilationLength contractionMass change

Book: pg. 66-67, 110-111,190-191

Test will cover :Relativity/Special relativity E=mc2

Work: W=FdKinetic Energy: KE= ½ mv2

Work-KE Theorem : W= ΔKEPotential Energy: PEg= mgh

Elastic potential Energy: PEelastic= ½ kΔx2

Conservation of Energy: KEi+ PEi = KEf+ PEf

Power: P= W/t = Fd/t = Fv = mad/t = mgd/t

Momentum: p = mvImpulse: Ft = Δp= mvf-mvi