chapter 5 work and energy. work and energy a force that causes a displacement of an object does work...
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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