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Work & Energy
WORK
Force x Distance
WORK
Force x Distance
Energy
Kinetic + (Potential)
Work and Energy
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Chapter 6 Roadmap
Method Differences
Work and energy
Crate example
2 important points about work
Work and Energy
• Combination of Force, Distance, and how they’re working together
creates scalar WORK.
• WORK either increases or decrease scalar KINETIC ENERGY –
involves velocity magnitude.
• Some types of WORK are always difference of 2 endpoints, and can
be treated as difference in scalar POTENTIAL ENERGY.
• LOSS OF PE often equals GAIN OF KE (or vice-versa). Thus
POTENTIAL + KINETIC (scalar) is CONSERVED
• Great shortcut – Solve complicated paths looking only at endpoints!
Method Differences
• Chapter 3
– Position, velocity, acceleration vectors.
– X and y components.
• Chapter 4
– Force and acceleration vectors.
– ΣF = ma is vector equation.
– Solve F=ma in x and y directions.
• Chapter 5
– Force and acceleration vectors.
– Solve F=ma in radial and other directions.
• Chapter 6
– Work and energy scalars.
– Forget direction, throw everything in “big mixing pot”.
Review Chapter 3
• From chapter 3
𝒗 = 𝒂𝑡 + 𝒗𝒐
𝒙=
1
𝒂𝑡 2
2
+ 𝒗𝒐 𝑡 + 𝒙𝒐
(solve for time)
(plug time in here)
• Combined to give
2𝒂 𝒙 − 𝒙𝒐 = 2𝒂 ∙ ∆𝒙 = 𝒗2 − 𝒗𝒐 2
• Required multiplication of vectors!
• Defined “scalar product”
– 𝑨 ∙ 𝑩 = 𝐴 𝐵 cos 𝜃
– Magnitude of each times “how much they’re inline”
• 𝒗 ∙ 𝒗 = 𝑣𝑣𝑐𝑜𝑠0 = 𝑣 2
• 𝒂 ∙ 𝒙 = 𝑎𝑥 cos 𝜃
(Boldface = vector)
Modifying 3rd Equation
• Modified 3rd Equation
2𝑎 ∆𝑥 cos 𝜃 = 𝑣 2 − 𝑣𝑜 2
• Consider several cases
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𝐼𝑓 𝒂 𝑖𝑛𝑙𝑖𝑛𝑒 𝑤𝑖𝑡ℎ ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒,
𝑣 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠
–
𝐼𝑓 𝒂 𝑎𝑙𝑚𝑜𝑠𝑡 𝑖𝑛𝑙𝑖𝑛𝑒 𝑤𝑖𝑡ℎ ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑠𝑜𝑚𝑒𝑤ℎ𝑎𝑡 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒,
–
𝐼𝑓 𝒂 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑧𝑒𝑟𝑜,
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𝐼𝑓 𝒂 𝑎𝑙𝑚𝑜𝑠𝑡 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑠𝑜𝑚𝑒𝑤ℎ𝑎𝑡 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒,
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𝐼𝑓 𝒂 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒,
𝑣 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎 𝑙𝑖𝑡𝑡𝑙𝑒
𝑣 𝑟𝑒𝑚𝑎𝑖𝑛𝑠 𝑠𝑎𝑚𝑒
𝑣 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎 𝑙𝑖𝑡𝑡𝑙𝑒
𝑣 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠
• Product of a and Δx, and how they’re working together, either
increases/decreases/keeps-constant v2
• Note v2 is scalar, no direction!
Work and Energy
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Modified 3rd Equation
2𝑎 ∆𝑥 cos 𝜃 = 𝑣 2 − 𝑣𝑜 2
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Multiply by ½ m
𝑚𝑎 ∆𝑥 cos 𝜃 =
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1
1
𝑚𝑣 2 − 𝑚𝑣𝑜 2
2
2
ma = Force
𝐹 ∆𝑥 cos 𝜃
1
= 𝑚𝑣 2 −
2
1
𝑚𝑣𝑜 2
2
𝑊𝑜𝑟𝑘 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 − 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙
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Work equals change in Kinetic Energy
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All scalars, use only magnitudes!
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Units N-m, or kg m2/s2 Joules (J)
Conclusions
• Product of force, distance, and how they’re working
together increases or decreases the magnitude of v.
• How force and distance work together is very important.
– If f and d inline, magnitude of v increases.
– If f and d partially inline, magnitude of v increases a little.
– If f and d perpendicular, magnitude of v remains constant.
– If f and d partially opposed, magnitude of v decreases a little.
– If f and d opposed, magnitude of v decreases.
– If f but no d v remains constant.
Work Definition
• Definition
F
F . x . cos(θ)
x
• Cos(θ) extracts F and x working together
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+1 when together
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-1 when opposed
-1 to +1 when in between
0 when perpendicular
• Work is a scalar quantity
Work done by Crate
• Example 6.1
– 50 kg crate, pulled 40 m
– FP = 100 N, Ffric = 50 N
• Method 1
– Solve for net force
– 100 N cos(37) – 50 N = 30 N
– Multiply by 40 m = 1200 J
• Method 2
– Find individual works
– Wmg = 0, WFn = 0, WFP = 3200, WFfric = -2000
– 0J + 0 J + 3200 J – 2000 J = 1200 J
• Work of sum = sum of works
Problem 8 Man lowering piano
• Forces
– Fg = 3234 N
– Ffric = μ mg cosθ = 1142 N
– FP = mg sinθ - μ mg cosθ = 376 N
• Works
– Wfr = 1142 N x 3.6 m (-1) = -4111 J
– WP = 376 N x 3.6 m
(-1) = -1353 J
– Wg = 3234 N x (3.6 sin28) = +5465 J
– Wnormal = 0 (perpendicular)
• Total work is 0
• Work of gravity was Fg times height
• Had it accelerated work would not be 0
Problem 8 – Work done by gravity
3234
3.6 sin28
3234 sin28
3.6
• Work done by gravity
– Force component along incline times total incline distance. 3234 sin 28 ∙ 3.6
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or
– Distance component along vertical times total vertical force. 3234 ∙ 3.6 sin 28
• 2nd is just weight times height (mgh)
Two important things
• Total Work is
– The work of the sum of all forces ΣFi x distance
• or
– The sum of the individual works of all forces. Σ(Fi x distancei)
• Individual Work is
– Force component in direction of displacement.
• or
– Displacement component in direction of force.
Work and Energy
Fx cosϴ = ½ mv2 - ½ mvo2
Work = ΔEnergy
Work equals change in energy
Examples of Work and Energy
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Example 6.5 – Work to increase car speed
Problem 18 – Work to stop car
Problem 23 - Air resistance on baseball
Example 6.8 – Falling baseball
– Use 2nd law
– Use work
• Example 6-9 - Roller coaster
– Use work
– Couldn’t do easily by 2nd law!
• Vertical circle example (use work)
• Note how you “mix up” dimensions!
Outta here
October Potter County hiking / camping