Title: Work, Power, Kinetic and Potential Energy sec 8'1 8'5 pg' 103108
1- 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
4- 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
5Conservation 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
6- 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
7- Conservation of Energy (Sec 8.6 pg. 109-111 pg.
117) - TEo TEf TE KE PE PE mgh KE ½ mv2
8Conservation 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
9b. 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