Chapter+2+Section+8+-+End

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toc =Section 8=

What Do You Think?
1/14/2011 There is not enough restoring force in the 11 m long pole to make the person go 12 m up. The factors that affect the height available to be attained is the height and elasticity of the pole, and of course, the ability of the pole vaulter himself.

Investigate
1a. The technique using our ruler to put the penny into the air 1b. The restoring force is the same as the distance the ruler bends * the spring constant. 5a. More deflection depending on how far the ball was from the ruler 5b There will be more bend the faster the vaulter runs

The less weight and the more bend there is, the higher the object will be propelled. In other words, the more the bend is and the less the mass is, the more force is exerted.

Physics Talk
1/18/10

Kinetic energy is the energy associated with motion KE=1/2MV^2. Gravitational potential energy is the energy an object possesses because of its vertical positioning compared to earth (GPE = mgh) Potential energy is the energy with respect to position The law of conservation of energy is energy that cannot be created or destroyed; it can be transformed from one form to another, but the total amount of energy remains constant Work is the product of the displacement and the force in the direction of the displacement work=f x d Elastic potential energy is the energy of a spring EPE=1/2 kx^2 k is the spring constant x is the distance of compression

Checking Up
1/18/10

1. A force is required to change the energy of an object 2. From the EPE 3. The pole vaulter gets his energy that bends the pole from his sprint 4. Joules are the units for KE, GPE, EPE

Class Notes
1/18/2010 Work = the product of force on an object and the distance the object moves because of this force. w = f*d ex: A 25kg object is lifted - how much work is done on the object if we lift it 1.5m? w = f*d w = fg * d w = (mg) *d w = (25g * 9.8) * 1.5 w = 367.5 joules. A joule is a measure of work and a measure of energy. This means that work and energy are equivalent. Energy is used to do work, and work is done to produce energy. Work = energy. Distance must be in the same direction as the force

EPE = elastic potential energy EPE = 1/2 Kx^2, x = is the amount of stretch or bend in meters. k is the spring constant in Newtons/meter <--- energy in a spring k = EPE *2 / x^2 F = Kx < --- force in a spring

GPE = gravitational potential energy GPE = mgh, m = mass, g = gravity, h = height in meters

KE = kinetic energy ME = 1/2 mv^2, m = mass in kg, v = velocity in m/s

Active Physics Plus
1/19/2011

Sample Problem B
The spring gives the toy KE which then gives it GPE. The entire process: Someone does work on a spring -> EPE -> converts to -> KE -> converts to -> GPE

B. KE = 1/2(mv^2) GPE came from KE so KE = GPE 1/2mv^2 = 118J m = .100kg. 1/2 (.1)(v^2) = 1.18 v = 4.9 m/s

C. x = 2cm = .02m k = ? EPE = 1/2kx^2 EPE = 1.18 1.18 = 1/2 k (0.2)^2 1.18 = 1/2 k(.0004) k = 5900 N/m

D. F = ? F = -kx F = 5900N/m(-.02) F = 118N

Physics To Go:
1/21/2011 1. You have to do work on shot put, it has GPE, and does more work when you throw it

2. Apply force to ball, do work and push the ball, and give the ball KE

3. KE=GPE 1/2mv^2=mgh 1/2v^2=gh 1/2(12)^2=(9.8)h h=7.35m

4. vaulting height is determined by speed and skill

5. It gets warmer because of energy. Whenever you bend it it breaks a little bit.

6. GPE=KE mgh=1/2mv^2 gh=1/2v^2 v=9.4m/s

7. gh=1/2v^2 v= 11m/s

8a. GPE=KE mgh=1/2mv^2 gh=1/2v^2 v= sqrt[(2)(9.8)(100)] v= 44m/s 8b. yes. This implies all objects fall at the same rate

9a. x=25cm= .25m k=1500 N/m W=? W=EPE W=1/2kx^2 W=1/2(1500N/m)(.25)^2 W=47J 9b. KE=W=47J 1/2mv^2=47J 1/2(.1)(v^2)=47 V= 31m/s

10a. k=315N/m w=? w=EPe=1/2kx^2 w=1/2(315)(.30)^2 w=14J 10b. F=kx F=(315)(.30) F=95N

11. GPE=KE=EPE mgh=1/2kx^2 (.04)(9.8)(1m)=1/2(18)x^2 x= .2m

12a. F=ma 1N=1kgm/s^2 F=ma 1(kg)(m/s^2) 12b. GPE=mgh GPE= (kg)(m/s^2)(m) 1J= 1Nm 12c. KE=1/2mv^2 J= kg(m/s)^2 J=1Nm 12c. EPE=1/2kx^2 1J=(N/m)(m)^2 1J=1Nm

13. diver uses KE running, jumps using GPE which transfers to KE, the EPE on the diving board, then KE, then GPE then KE again and they hit the water

14. KE when it comes to her, compress EPE, back to KE, GPE, then the teamate hits it with EPE, Ke over the net GPE is lost as KE increases

=Section 9=

What Do You See?
1/31/2011 There is a skater in the air spinning around while the lady in the helicopter is timing him

What Do You Think?
1/31/2011 I think that the hang time of some athletes do not defy the pull of gravity, unless, of course, in the presence of no gravity. Gravity always pulls objects down towards the earth, and the hang time can't defy that unless the athlete is not on the earth. The skater does not defy gravity because it pulls him down anyways. The force of gravity is acting on the skater so he cannot possibly defy it.

Investigate: Analysis of the Triple Axel
1/31/2011 1a. 20 times 1b. 1 frame = 1/30 seconds. 20 frames = 20(1/30) = 20/30 seconds. 2/3 seconds. 1c. Yes, the skater looked like he was floating in the air. But realistically, he obviously was not.

2a. Basketball player = 31 frames. 31 * (1/30) = 31/30 seconds. 2b. Yes, the player hung because hes in the air for a time.

3.Weight(N) = Weight(lbs) * 4.38 N/lb Weight(N) = 147 LBs * 4.38 N/lb Weight(N) = 643.86N

Weight(N) = mg 643.86 = m (9.8) m = 65.7 kg

4a. 0.875 meters

5a.1.065 meters

5b.1.065 - 0.875 = 0.19 meters

6a. 1.31 meters

6b. 1.31 - 1.065 = 0.245 meters

7a. GPE = mgh GPE = 68 kg (9.8) (.245) GPE = 163.268 J

7b. EPE = GPE EPE = 163.268 J

7c. GPE = mgh GPE = 68 (9.8) (.19) GPE = 126.616 J

Physics Talk
2/1/2011 When you are jumping, and when you are in the ready position you have elastic potential energy(EPE). Potential energy is present due to chemical reactions in muscles. As you move EPE exchanges to GPE/KE. Conservation of energy can be described as EPE, KE, GPE, sound energy, light energy, chemical energy, electrical energy, nuclear energy, and internal energy

Checking Up
2/1/2011 1. Energy in the launch position comes from chemical reaction in the muscles 2. In the launch position the student will have an increase in GPE and KE. In the peak of the jump, they will lose KE and gain GPE. 3. Sound energy, light energy, chemical energy, electrical energy, nuclear energy, and internal energy

Physics Plus
2/3/2011 Analysis of a Roller Coaster

m = 50 g = 0.050kg Calculate GPE, KE, Etotal, V at A, B, C, and D Ignore all forms of friction(includes air resistance) This means energy is conserved Energy is not dissipated.

Etotal is constant.

A. GPE = mgh GPE = .050(9.8)(.45m) GPE = 0.22J KE = 1/2(m)v^2 KE = 1/2(0.050)(0.0)^2 KE = 0 Etotal = 0.22J V = 0

B. GPE = mgh GPE = 0.050(9.8)(0) GPE = 0 KE= Et - GPE KE = 0.22J - 0 KE = 0.22J Etotal =0.22J V = 2.9 m/s

C. GPE = mgh GPE = 0.050(9.8)(0.28) GPE = 0.137 J KE= Et - GPE KE = 0.22J - 0.137 KE = 0.08J

Etotal = 0.22J V =1.8m/s

D. GPE = mgh GPE = 0.050(9.8)(0) GPE = 0 KE = 1/2(m)v^2 KE = 1/2(0.050)( KE = Etotal = 0.22J V =

Physics To Go
2/4/2011 1. W = F * D F = mg F = 50KG(9.8) W = 490 * 1M w = 490J

2. You need work to start the sled, but before it starts it has no energy. As work is applied, kinetic energy is obtained, and as the sled starts accelerating, more energy is being outputted. When the sled is stopped, there is no more energy and no more work.

3. There is no possible way that a person can hang in the air on their own accord without the use of supportive equipment. The person looked as to hang in the air for a second because he reached his apogee, and had a velocity of close to 0 for longer than he had a velocity that was accelerating.

4. The person needs to provide empirical evidence as to how their proof is true. If something is taken as true without being proved false, then ideas such as the world being flat would still have prevailed if not for a person exploring the situation even more.

5. An athlete can increase his or her vertical jump height by losing weight and gaining muscle to exert more force, therefore getting more work and more energy.

6. a. W = F * D F = mg F = 1N(9.8) W = 1 * 1M w = 1J

b. W = F * D F = mg F = 1N(9.8) W = 1 * 10M w = 10J

c. W = F * D F = mg F = 10N(9.8) W = 10 * 1M w = 10J

d. W = F * D F = mg F = 100N(9.8) W = 100 * 0.1M w = 10J

7. a. GPE = mgh GPE = (1N)(9.8)(1) GPE = 1J

b. GPE = mgh GPE = (1N)(9.8)(10) GPE = 10J

c. GPE = mgh GPE = (10N)(9.8)(1) GPE = 10J

d. GPE = mgh GPE = (100N)(9.8)(.1) GPE = 10J

8. KE = 1/2mv^2 KE = 1/2(0.101)(

9. W = F*D W = 50N * 43m W = 2,150 J

10. KE = 1/2 mv^2 KE = 1/2 (62 kg) (8.2m/s)^2 KE = 254.2 J

11a. F = ma 30 N = 5(a) a = 6 m/s^2

11b. W = f*d W = 30*18.75 W = 562.5J

12a. w = f*d 40000N = 3200 * d d = 12.5

12b. F = ma 3200N = 1200K *a a = 2.667m/s^2

13. KE = 1/2 mv^2 KE = 1/2 (.15kg)(40m/s)^2 KE = 120J

14. F = ma 417 = 64Kg a a = 6.515m/s^2

6.515 = (15-0)/t t = 2.3 seconds

d = 1/2(at^2) d = 1/2 (6.515)(2.3)^2 d = 17.232