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Work & Energy WORK Force x Distance WORK Force x Distance Energy Kinetic + (Potential) Work and Energy • • • • • • Intro example Chapter 6 Roadmap Method Differences Work and energy Crate example 2 important points about work Chapter 2 review • Start with equation 2-11c 𝑣 2 = 𝑣𝑜 2 + 2𝑎𝑥 • v2 and vo2 are scalars e.g. 2𝑚 𝑠 2 = −2 𝑚 𝑠 2 • So 2ax must be scalar – If a and x in same direction, v2 gets larger – If a and x in opposite direction, v2 gets smaller – If a and x perpendicular, v2 stays same 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, v2 gets larger – If a and x in opposite direction, 2ax negative, v2 gets smaller – If a and x perpendicular, 2ax zero, v2 stays same Work and energy • Equation 2-11c 𝑣 2 = 𝑣𝑜 2 + 2𝑎𝑥 • Multiply by ½ m 2 1 1 𝑚𝑣 2 = 𝑚𝑣𝑜 + 𝑚𝑎𝑥 2 2 2 1 1 𝑚𝑣 2 = 𝑚𝑣𝑜 + 𝐹𝑥𝑐𝑜𝑠𝜃 2 2 New kinetic energy Old kinetic energy Work Work and Energy 𝐹 ∆𝑥 cos 𝜃 1 = 𝑚𝑣 2 − 2 1 𝑚𝑣𝑜 2 2 𝑊𝑜𝑟𝑘 = 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦𝑓𝑖𝑛𝑎𝑙 − 𝐾𝑖𝑛𝑒𝑡𝑖𝑐 𝐸𝑛𝑒𝑟𝑔𝑦𝑖𝑛𝑖𝑡𝑖𝑎𝑙 • 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 − 𝑣𝑜 2 • Consider several cases – 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, – 𝐼𝑓 𝒂 𝑎𝑙𝑚𝑜𝑠𝑡 𝑜𝑝𝑝𝑜𝑠𝑖𝑡𝑒 ∆𝒙, 𝑙𝑒𝑓𝑡 𝑠𝑖𝑑𝑒 𝑠𝑜𝑚𝑒𝑤ℎ𝑎𝑡 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒, – 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! 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 – 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”. 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 • Definition F F . x . cos(θ) x • 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 Work done by Crate (ex 6-7) • 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 Work done by gravity -1 • Force component in direction of displacement • Using angle between force and displacement - ϴ 𝑊𝑜𝑟𝑘 = mgcos 𝜃 ∙ 𝑑 𝑊𝑜𝑟𝑘 = dcos 𝜃 ∙mg ϴ h mg cosϴ 𝑊𝑜𝑟𝑘 = d𝑠𝑖𝑛 𝜑 ∙mg displacement d 𝑊𝑜𝑟𝑘 = ℎ ∙mg mg φ Work done by gravity - 2 • Force component in direction of displacement • Using complimentary angle Φ 𝑊𝑜𝑟𝑘 = mg𝑠𝑖𝑛(𝜙) ∙ 𝑑 mg sinΦ displacement mg Φ Φ Work done by gravity - 3 • Displacement component in direction of force • Another way of looking at sin(Φ) 𝑊𝑜𝑟𝑘 = mg ∙ 𝑑𝑠𝑖𝑛(𝜙) Same thing! d sinΦ Φ displacement mg Φ So the work done by gravity is just mgh Man lowering piano – Prob 10 • 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 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) • Each Individual Work – Force component in direction of displacement. • or – Displacement component in direction of force. Work and Energy 1 1 2 𝐹 ∙ 𝑥 cos 𝜃 = 𝑚𝑣𝑓 − 𝑚𝑣𝑖 2 2 2 𝑊𝑜𝑟𝑘 = ∆𝐾𝐸 Work equals change in Kinetic 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 – Use 2nd law – Use work • Example 6-8 - Roller coaster – Use work – 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 Outta here October Potter County hiking / camping