Work, Power, Kinetic and Potential Energy sec 8'1 8'5 pg' 103108

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• Work, Power, Kinetic and Potential Energy (sec
  8.1 8.5 pg. 103-108)
• Work Work is the product of force exerted on an
  object in the direction of motion and the
  _______________ the object travels. Work
  corresponds to a change in energy.
• The SI unit of work is the Joule (J). 1 J 1
  Nm.
• w F d
• Examples
• Burt has a mass of 125 kg and stands on a pogo
  stick. When Burt stands on a pogo stick, the
  spring compresses 0.050 m. How much work is
  performed on the spring?
• w Fd (the force impressed upon
  the spring is Burts weight)
• w mgd
• w (125 kg)(9.80 m/s2)(0.050 m)
• w 61 J
• Sandy has a mass of 50.0 kg and climbs 2.0 m.
  How much work was done during the climb? What was
  the change in potential energy?
• wFd (the force Sandy must supply is
  to overcome her weight)
• wmgd
• w (50.0 kg)(9.80 m/s2)(2.0 m)
• w 980 J
• 3. A block is pushed 2.5 m by a net force of
  50.0 N in the direction of motion. How much work
  was done?
• wFd
• w (50.0 N)(2.5 m)
• w 130 J

distance
?PE w
2

• Power Power the rate at which work is performed
  on some object. Therefore, power is the quotient
  of work and time.
• P w/t
• The SI unit of power is the __________ (W). 1
  W 1 J/s
• A Joule per second is a Watt?
• Example 4 From example 2, calculate the power
  generated by Sandy if she performed the work in
  2.1 s.
• P w / t
• P 980.0 J / 2.1 s
• P 470 W
• Energy Energy is the ability to do work. There
  are many forms of energy we will study 2 of
  those forms (kinetic and potential energy).
  Energy is also measured in Joules (J).
• Potential Energy the energy an object possesses
  due to its position (a stored energy). When an
  object is lifted to higher levels (such as
  climbing a ladder), there is an _____________ in
  potential energy. When you pull back on a rubber
  band, there is an increase in potential energy.
  Potential energy is a product of force and
  displacement (remember work corresponded to a
  change in energy).
• PE Fd (when looking at gravitation acting on
  a body F mg)
• for objects at some height above the ground we
  will use PE mgh

Watt
Exactly
Petersen factor
increase
3

• Example 5 a. A 0.20 kg apple hangs 7.0 m above
  the ground from a tree. Calculate the potential
  energy of the apple.
• PE mgh
• PE (0.20 kg)(9.80 m/s2) (7.0 m)
• PE 14 J
• What will be the change in potential energy if
  the apple falls and lands on a table located 2.0
  m above the ground?
• ?PE PEf PEo PEf mghf
• PEf 0.20 kg(9.80 m/s2)(2.0m)
• PEf 3.920 J
• How will the PE lost during the fall compare to
  the KE gained during the fall?
• Example 6 Billy pulls back on a rock in his
  slingshot a distance of 5.0 cm by pulling with a
  force of 100. N. How much work was done on the
  rock and what is the change in potential energy
  of the rock?

b. ? PE PEf PEo ? PE mghf mgho
? PE mg (hf ho) ? PE (0.20 kg)(9.80
m/s2) (2.0 m 7.0 m) ? PE (0.20 kg)(9.80
m/s2) ( 5.0 m) ? PE -9.8 J
?PE 3.920 J - 13.72 J
?PE 10. J
The negative sign indicates a loss in energy of
the apple
The amount of potential energy lost by the apple
will be equal to the kinetic energy gained by the
apple. Or ?PE - ?KE
The amount of potential energy lost by the apple
will be equal to the kinetic energy gained by the
apple. Or ?PE - ?KE
w Fd
w (100. N)(5.0 cm (m / 100 cm))
?PE w 5.0 J
w 5.0 J
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• Kinetic Energy the energy an object possesses
  due to its motion. Kinetic energy varies
  linearly with mass and with the __________ of the
  velocity of an object.
• KE ½ mv2
• Example 7 Calculate the KE of
• a. A 10.0 kg ball traveling 5.00 m/s
• KE ½ mv2
• KE ½ (10.0 kg)(5.00 m/s)2
• KE 125 J
• Example 8 If a 50. kg cart traveling along a
  horizontal surface 10.0 m/s slides to a halta.
  What is the change in kinetic energy of the cart?
  
• b. How much work is performed on the cart by the
  horizontal surface?
• c. If the cart is brought to a halt over a
  distance of 5.00m, what is the frictional force
  acting on the cart?

square
b. a 5.00 kg ball traveling 10.0 m/s
KE ½ mv2
KE ½ (5.00 kg)(10.0 m/s)2
KE 250. J
?KE KEf - KEo
/
?KE - ½ mvo2
?KE - ½ (50. kg)(10.0 m/s)2
?KE -2500 J
w ?KE or -2500 J
F w/d
F -2500 J / 5.00 m
F -5.0.102 N
The negative sign indicates that the force
opposes the motion of the object
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Conservation of Energy
A 500. kg rock is pushed off of a cliff that is
200. m tall

• 1a. Determine the potential energy of the rock at
  the top of the cliff.
• PE mgh
• PE (500. kg)(9.80 m/s2)(200. m)
• PE 9.80.105 J
• b. What is the KE of the rock while resting at
  the top of the cliff?
• KE at the top is 0 J
• c. What is the total energy of the rock at the
  top of the cliff?
• TE KE PE 9.80.105 J
• d. At the bottom of the cliff, what will be the
  PE?
• PE at bottom 0 J, KE at bottom is 9.80.105 J
• e. What will be the velocity of the rock at the
  bottom of the cliff?
• KEf ½ mv2
• v (2KE / m)1/2
• v (2 (9.800 .105 J) / 500. kg)1/2

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• f. What will be the values of the KE and PE of
  the rock when it is falling at a speed of 20.
  m/s?
• KE ½ mv2
• KE ½ (500. kg)(20. m/s)2
• KE 1.0.105 J
• PE TE KE
• PE 9.800 .105 J 1.00.105 J
• PE 8.8.105 J

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• Conservation of Energy (Sec 8.6 pg. 109-111 pg.
  117)
• TEo TEf TE KE PE PE mgh KE ½ mv2

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Conservation of Energy continued
a. 200. kg rock is pushed off of a 200. m cliff.
What will be the height of the rock when it
falls at a rate of 10.0 m/s?

• 2a. hf ? when vf 10.0 m/s
• TE o TEf
• KEo PEo KEf PEf
• PEf PEo KE f
• mghf mgho ½ mvf2
• hf (gho ½ vf2 )
• g
• hf (9.80 m/s2(200. m)) ½ (10.0 m/s)2
• 9.80 m/s2
• hf 1960 m2/s2 50.00 m2/s2
• 9.80 m/s2
• hf 1910 m2/s2 / 9.80 m/s2
• hf 195 m

You could solve for each (shown here) or
substitute equations (shown left) PEo mgho PEo
200. kg(9.80 m/s2)(200. m) PEo 392000 J
/
KEf ½ mvf2 KEf ½ (200. kg)(10.0 m/s)2 KEf
10000 J
PEf PEo KEf PEf 392000 J 10000 J mghf
382000 J hf 382000 J / mg hf 382000 J /
(200.kg(9.80 m/s2)) hf 195 m
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b. If the rock were thrown downward from the same
cliff with an initial velocity of 10. m/s, what
will be the velocity of the rock when it is 100.
m above the ground?

• 2b. vf ? when vo 10. m/s and hf 100. m
• TE o TEf
• PEo KEo PEf KEf
• KEf PEo KEo PE f
• ½ mvf 2 mgho ½ mvo2 - mghf
• ½ vf2 ½ vo2 gho - ghf
• vf (2g(ho - hf) vo2)1/2
• vf (2 (9.80 m/s2)(200. m -100. m) (10.
  m/s)2)1/2
• vf (2 (9.80 m/s2)(100. m) (10. m/s)2)1/2
• vf (1960 m2 /s2 100 m2/s2)1/2
• vf (2060 m2 /s2)1/2
• vf 45.4 m/s
• vf (2g(ho - hf) vo2)1/2
• is the same as
• vf (2ad vo2)1/2

You could solve for each (shown here) or
substitute equations (shown left) PEo
mgho PEo 200. kg(9.80 m/s2)(200. m) PEo
392000 J KEo ½ mvo2 KEo ½ (200. kg)(10.
m/s)2 KEo 10000 J PEf mghf PEf (200.
kg)(9.80m/s2)(100.m) PEf 196000 J KEf PEo
KEo PEf KEf 392000 J 10000 J -196000 J KEf
206000 J KEf ½ mvf2 vf (2KEf / m)1/2 vf
(2(206000 J) / 200. kg)1/2 Vf 45.4 m/s