Transcript Slide 1

Gravitational Potential Energy
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Review Work and Energy
Work due to Gravity
Writing as Potential Energy difference
Examples
Problem 8 – Work done by gravity
Note: use sin rather
than cos for angle
from horizontal
Forces
3234
3234 sin28
Displacements
3.6 sin28
3.6
• Work done by gravity
– Force component along incline times total incline distance. 3234 sin 28 ∙ 3.6
•
or
– Distance component along vertical times total vertical force. 3234 ∙ 3.6 sin 28
• 2nd is just weight times height (mgh)
Work done by gravity
• Work done my gravity
This is the
path the
ball takes
But this is the path
you can use for
calculating the
work!
• And the string tension does no work!
Work done by Gravity
• In both cases work is:
𝑊𝑜𝑟𝑘𝑔𝑟𝑎𝑣 = 𝑡𝑜𝑡𝑎𝑙 𝑓𝑜𝑟𝑐𝑒 𝑜𝑓 𝑔𝑟𝑎𝑣𝑖𝑡𝑦 ∙ ℎ𝑒𝑖𝑔ℎ𝑡 𝑐ℎ𝑎𝑛𝑔𝑒
𝑊 = 𝑚𝑔ℎ
• If object starts high at yi and falls low to yf work is
𝑊 = 𝑚𝑔(𝑦𝑖 − 𝑦𝑓 )
– Only depends of height difference
– Is positive for high to low.
– Is negative for low to high.
• Thus work done by gravity can always be written
𝑊 = 𝑚𝑔𝑦𝑖 − 𝑚𝑔𝑦𝑓 = − 𝑚𝑔𝑦𝑓 − 𝑚𝑔𝑦𝑖
Looks like a difference of two “energies”!
Potential Energy
• Work done by gravity
𝑊 = −(𝑚𝑔𝑦𝑓 − 𝑚𝑔𝑦𝑖 )
𝑊𝑜𝑟𝑘 = 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 − 𝑃𝑜𝑡𝑒𝑛𝑡𝑖𝑎𝑙 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙
• Compare with work vs. Kinetic Energy
𝑊=
1
𝑚𝑣𝑓 2 −
2
1
𝑚𝑣𝑖 2
2
𝑊𝑜𝑟𝑘 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 − 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙
• Thus loss of Potential Energy = Gain of Kinetic Energy
− 𝑚𝑔𝑦𝑓 − 𝑚𝑔𝑦𝑖
= 𝑊𝑜𝑟𝑘 =
1
𝑚𝑣𝑓 2 −
2
1
𝑚𝑣𝑖 2
2
Conservation of KE + PE
• Loss of Potential Energy = Gain of Kinetic Energy
− 𝑚𝑔𝑦𝑓 − 𝑚𝑔𝑦𝑖
= 𝑊𝑜𝑟𝑘 =
1
𝑚𝑣𝑓 2 −
2
1
𝑚𝑣𝑖 2
2
• Rearranging
1
1
2
𝑚𝑣𝑖 + 𝑚𝑔𝑦𝑖 = 𝑚𝑣𝑓 2 + 𝑚𝑔𝑦𝑓
2
2
• Result
If only gravity is acting, or gravity plus forces that do no work,
then sum of kinetic + gravitational potential energy is conserved
Potential Energy
• Without potential energy
• With potential energy
KE+PE
KE+PE
KE+PE
• Do not do both, you’ll be double-counting!
KE+PE
Examples of Potential Energy
• Example 6-8 (falling rock)
• Example 6-9 (roller coaster)
• Example - Vertical circle
Examples of Potential Energy I
• Rock falling 3 m – Work/Energy
= 𝑚𝑣𝑜 + 𝑓𝑜𝑟𝑐𝑒 ∙ 𝑑𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑐𝑜𝑠𝜃
1
𝑚𝑣 2
2
= 𝑚𝑔 ∙ 𝑦 𝑐𝑜𝑠0
1
𝑚𝑣 2
2
= 𝑚𝑔 ∙ 𝑦
v=
1
2
2
1
𝑚𝑣 2
2
2 ∙ 9.8 𝑚 𝑠 2 ∙ 3𝑚
𝑣 = 7.7 𝑚/𝑠
• Rock falling 3 m – Conservation of Energy
1
2
𝐸𝑖 = 𝑚𝑣𝑖 2 + 𝑚𝑔ℎ𝑖
1
2
𝐸𝑓 = 𝑚𝑣𝑓 2 + 𝑚𝑔ℎ𝑓
1
𝑚𝑣𝑓 2
2
v=
= 𝑚𝑔ℎ𝑖
2 ∙ 9.8 𝑚 𝑠 2 ∙ 3𝑚 = 7.7 m/s
Examples of Potential Energy II
• Example 6-9 - Roller coaster
– Forces mg, normal force
– Normal force does no work!
vo
– Initial Energy
1
2
• 𝐸𝑖 = 𝑚𝑣𝑖 2 + 𝑚𝑔ℎ𝑖
– Final Energy (bottom)
1
2
• 𝐸𝑓 = 𝑚𝑣𝑓 2 + 𝑚𝑔ℎ𝑓
– Conservation of Energy
•
1
𝑚𝑣𝑓 2
2
= 𝑚𝑔ℎ𝑖
v
Examples of Potential Energy III
• Example 6-9 - Roller coaster
– Forces mg, normal force
– Normal force does no work!
vo
– Initial Energy
1
2
• 𝐸𝑖 = 𝑚𝑣𝑖 2 + 𝑚𝑔40
– Final Energy (other side)
v
1
2
• 𝐸𝑓 = 𝑚𝑣𝑓 2 + 𝑚𝑔25
– Conservation of Energy
•
1
𝑚𝑣𝑓 2
2
= 𝑚𝑔40 − 𝑚𝑔25
» Don’t even need to consider the middle position!
Examples of Potential Energy IV
• Vertical circle
– Forces mg, Tension
– Circle radius r
– Tension does no work!
– Initial Energy
1
2
• 𝐸𝑖 = 𝑚𝑣𝑖 2 + 𝑚𝑔2r
– Final Energy
1
2
• 𝐸𝑓 = 𝑚𝑣𝑓 2 + 𝑚𝑔ℎ𝑓
– Conservation of energy
•
1
𝑚𝑣𝑓 2
2
1
2
= 𝑚𝑣𝑖 2 + 𝑚𝑔2𝑟
Example of Potential Energy V
• Problem 3-31
– Compare with Chapter 3
– Must include velocity at maximum
– f) Max Height (202 m, 78 above cliff)
– d) Velocity at ground (81.7 m/s)
Example of Potential Energy VI
• Problem 40
Problem 40
• Critical point as ball slows down
• 2 basic equations:
– 𝑚𝑔 =
𝑚𝑣 2
𝑟
– 𝑚𝑔ℎ = 𝑚𝑔2𝑟 +
(circular)
1
𝑚𝑣 2
2
(energy)
• Devious trick
–
1
𝑚𝑣 2
2
– 𝑚𝑔ℎ =
5
2
1
2
= 𝑚𝑔𝑟
– ℎ= 𝑟
5
𝑚𝑔𝑟
2
𝑟
2
(multiply 1st equation by )
1
2
(substitute for 𝑚𝑣 2 in second)
Atwood’s machine
• From chapter 3
𝑎=
𝑚2 −𝑚1
𝑔
𝑚2 +𝑚1
𝑣 2 = 2𝑎ℎ = 2
𝑚2 −𝑚1
𝑔ℎ
𝑚2 +𝑚1
• Initial Energy (m2 up m1 down)
1
2
1
2
𝐸𝑖 = 𝑚1 𝑣𝑖 2 + 𝑚2 𝑣𝑖 2 + 𝑚2 𝑔ℎ
• Final Energy (m1 up m2 down)
1
2
1
2
𝐸𝑖 = 𝑚1 𝑣𝑓 2 + 𝑚2 𝑣𝑓 2 + 𝑚1 𝑔ℎ
• Equating Initial and Final
1
2
2
𝑚1 + 𝑚2 𝑣𝑓 = 𝑚2 −𝑚1 𝑔ℎ
𝑣𝑓 2 = 2
𝑚2 −𝑚1
𝑔ℎ
𝑚2 +𝑚1
Same thing!
3 things to remember
• Work (or ΔPE) allows you to “mix up”
dimensions.
• Calculate Work or ΔPE, but not both.
• Zero point in gravitational PE arbitrary.