Work & Energy WORK Force x Distance WORK Force x Distance Energy Kinetic + (Potential)

Work and Energy Intro example Chapter 6 Roadmap Method Differences Crate example 2 important points about work

Chapter 6 review Start with equation 2-11c 𝑣 2 = 𝑣 𝑜 2 +2𝑎𝑥 𝑣 2 = 𝑣 𝑜 2 +2𝑎𝑥 v2 and vo2 are scalars e.g. 2 𝑚 𝑠 2 = −2 𝑚 𝑠 2 So 2ax must be scalar If a and x in same direction, larger v2 If a and x in opposite direction, smaller v2 If a and x perpendicular, same v2

Scalar product of 2 vectors Could define scalar product of a and x 𝑎∙𝑥= 𝑎 𝑥 𝑐𝑜𝑠𝜃 Result If a and x in same direction, 2ax positive, larger v2 If a and x in opposite direction, 2ax negative, smaller v2 If a and x perpendicular, 2ax zero, same v2

Work and energy Equation 2-11c Multiply by ½ m 𝑣 2 = 𝑣 𝑜 2 +2𝑎𝑥 𝑣 2 = 𝑣 𝑜 2 +2𝑎𝑥 Multiply by ½ m 1 2 𝑚 𝑣 2 = 1 2 𝑚𝑣 𝑜 2 +𝑚𝑎𝑥 1 2 𝑚 𝑣 2 = 1 2 𝑚𝑣 𝑜 2 +𝐹𝑥𝑐𝑜𝑠𝜃 New kinetic energy Old kinetic energy Work

𝑊𝑜𝑟𝑘 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑓𝑖𝑛𝑎𝑙 − 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 Work and Energy 𝐹 ∆𝑥 cos 𝜃 = 1 2 𝑚𝑣 2 − 1 2 𝑚 𝑣 𝑜 2 𝑊𝑜𝑟𝑘 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑓𝑖𝑛𝑎𝑙 − 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦 𝑖𝑛𝑖𝑡𝑖𝑎𝑙 Work equals change in Kinetic Energy All scalars, use only magnitudes! Units N-m, or kg m2/s2 Joules (J)

Generalizing equation 2-11c Modified 3rd Equation 2𝑎 ∆𝑥 cos 𝜃 = 𝑣 2 − 𝑣 𝑜 2 Consider several cases 𝐼𝑓 𝒂 𝑖𝑛𝑙𝑖𝑛𝑒 𝑤𝑖𝑡ℎ ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒, 𝑣 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝐼𝑓 𝒂 𝑎𝑙𝑚𝑜𝑠𝑡 𝑖𝑛𝑙𝑖𝑛𝑒 𝑤𝑖𝑡ℎ ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑠𝑜𝑚𝑒𝑤ℎ𝑎𝑡 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒, 𝑣 𝑖𝑛𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎 𝑙𝑖𝑡𝑡𝑙𝑒 𝐼𝑓 𝒂 𝑝𝑒𝑟𝑝𝑒𝑛𝑑𝑖𝑐𝑢𝑙𝑎𝑟 𝑡𝑜 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑧𝑒𝑟𝑜, 𝑣 𝑟𝑒𝑚𝑎𝑖𝑛𝑠 𝑠𝑎𝑚𝑒 𝐼𝑓 𝒂 𝑎𝑙𝑚𝑜𝑠𝑡 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑠𝑜𝑚𝑒𝑤ℎ𝑎𝑡 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒, 𝑣 𝑑𝑒𝑐𝑟𝑒𝑎𝑠𝑒𝑠 𝑎 𝑙𝑖𝑡𝑡𝑙𝑒 𝐼𝑓 𝒂 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒, 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!

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.

Method Differences Chapter 3 Chapter 4 Chapter 5 Chapter 6 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 Solve F=ma in radial and other directions. Chapter 6 Work and energy scalars. Forget direction, throw everything in “big mixing pot”.

Future Roadmap 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!

Work Definition F . x . cos(θ) +1 when together Definition Cos(θ) extracts F and x working together +1 when together -1 when opposed -1 to +1 when in between 0 when perpendicular Work is a scalar quantity F x

Work done by Crate Example 6.1 Method 1 Method 2 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 (1) Force component in direction of displacement Using angle between force and displacement - ϴ 𝑊𝑜𝑟𝑘=mgcos⁡(𝜃)∙𝑑 displacement mg ϴ mg cosϴ

Problem 8 – work done by gravity (2) Force component in direction of displacement Using complimentary angle Φ 𝑊𝑜𝑟𝑘=mg𝑠𝑖𝑛⁡(𝜙)∙𝑑 displacement mg Φ mg sinΦ

Problem 8 – work done by gravity (3) Displacement component in direction of force Another way of looking at sin(Φ) 𝑊𝑜𝑟𝑘=mg∙𝑑𝑠𝑖𝑛⁡(𝜙) Same thing! So the work done by gravity is just mgh displacement mg Φ d sinΦ

Two important things Total Work is Each Individual Work 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) Each Individual Work Force component in direction of displacement. Displacement component in direction of force.

Work equals change in Kinetic Energy Work and Energy 𝐹∙𝑥 cos 𝜃 = 1 2 𝑚 𝑣 𝑓 2 − 1 2 𝑚 𝑣 𝑖 2 𝑊𝑜𝑟𝑘=∆𝐾𝐸 Work equals change in Kinetic Energy

Work equals change in energy Work and Energy Fx cosϴ = ½ mv2 - ½ mvo2 Work = ΔEnergy Work equals change in energy

Examples of Work and Energy Example 6.7 – Falling rock Use 2nd law Use work Car going down ramp Example 6-8 - Roller coaster Couldn’t do easily by 2nd law! Vertical circle example (use work) Note how you “mix up” dimensions!

More Examples of Work and Energy Example 6.6 – Work to increase car speed Example 6.5 – Work to stop car Problem 23 - Air resistance on baseball

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