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Work & Energy WORK Force x Distance WORK Force x Distance Energy
Kinetic + (Potential)
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Work and Energy Intro example Chapter 6 Roadmap Method Differences
Crate example 2 important points about work
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Chapter 6 review Start with equation 2-11c π£ 2 = π£ π 2 +2ππ₯
π£ 2 = π£ π 2 +2ππ₯ v2 and vo2 are scalars e.g π π 2 = β2 π π 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
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Scalar product of 2 vectors
Could define scalar product of a and x πβπ₯= π π₯ πππ π 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
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Work and energy Equation 2-11c Multiply by Β½ m π£ 2 = π£ π 2 +2ππ₯
π£ 2 = π£ π 2 +2ππ₯ Multiply by Β½ m 1 2 π π£ 2 = ππ£ π 2 +πππ₯ 1 2 π π£ 2 = ππ£ π 2 +πΉπ₯πππ π New kinetic energy Old kinetic energy Work
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ππππ = πΎππππ‘ππ πΈπππππ¦ πππππ β πΎππππ‘ππ πΈπππππ¦ ππππ‘πππ
Work and Energy πΉ βπ₯ cos π = 1 2 ππ£ 2 β π π£ π 2 ππππ = πΎππππ‘ππ πΈπππππ¦ πππππ β πΎππππ‘ππ πΈπππππ¦ ππππ‘πππ Work equals change in Kinetic Energy All scalars, use only magnitudes! Units N-m, or kg m2/s2 Joules (J)
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Generalizing equation 2-11c
Modified 3rd Equation 2π βπ₯ cos π = π£ 2 β π£ π 2 Consider several cases πΌπ π ππππππ π€ππ‘β βπ, ππππ‘ π πππ πππ ππ‘ππ£π, π£ πππππππ ππ πΌπ π πππππ π‘ ππππππ π€ππ‘β βπ, ππππ‘ π πππ π ππππ€βππ‘ πππ ππ‘ππ£π, π£ πππππππ ππ π πππ‘π‘ππ πΌπ π ππππππππππ’πππ π‘π βπ, ππππ‘ π πππ π§πππ, π£ πππππππ π πππ πΌπ π πππππ π‘ πππππ ππ‘π βπ, ππππ‘ π πππ π ππππ€βππ‘ πππππ‘ππ£π, π£ πππππππ ππ π πππ‘π‘ππ πΌπ π πππππ ππ‘π βπ, ππππ‘ π πππ πππππ‘ππ£π, π£ πππππππ ππ Product of a and Ξx, and how theyβre working together, either increases/decreases/keeps-constant v2 Note v2 is scalar, no direction!
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Conclusions Product of force, distance, and how theyβre 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.
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Method Differences Chapter 3 Chapter 4 Chapter 5 Chapter 6
Position, velocity, acceleration vectors. X and y components. Chapter 4 Force and acceleration vectors. Ξ£F = 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 βbig mixing potβ.
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Future Roadmap Combination of Force, Distance, and how theyβre working together creates scalar WORK. WORK either increases or decrease scalar KINETIC ENERGY β 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 β Solve complicated paths looking only at endpoints!
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Work Definition F . x . cos(ΞΈ) +1 when together Definition
Cos(ΞΈ) 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
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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) β 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 J β 2000 J = 1200 J Work of sum = sum of works
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Problem 8 Man lowering piano
Forces Fg = 3234 N Ffric = ΞΌ mg cosΞΈ = 1142 N FP = mg sinΞΈ - ΞΌ mg cosΞΈ = 376 N Works Wfr = 1142 N x 3.6 m (-1) = J WP = 376 N x 3.6 m (-1) = J Wg = 3234 N x (3.6 sin28) = J Wnormal = 0 (perpendicular) Total work is 0 Work of gravity was Fg times height Had it accelerated work would not be 0
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Problem 8 β work done by gravity (1)
Force component in direction of displacement Using angle between force and displacement - Ο΄ ππππ=mgcosβ‘(π)βπ displacement mg Ο΄ mg cosΟ΄
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Problem 8 β work done by gravity (2)
Force component in direction of displacement Using complimentary angle Ξ¦ ππππ=mgπ ππβ‘(π)βπ displacement mg Ξ¦ mg sinΞ¦
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Problem 8 β work done by gravity (3)
Displacement component in direction of force Another way of looking at sin(Ξ¦) ππππ=mgβππ ππβ‘(π) Same thing! So the work done by gravity is just mgh displacement mg Ξ¦ d sinΞ¦
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Two important things Total Work is Each Individual Work
The work of the sum of all forces Ξ£Fi x distance or The sum of the individual works of all forces. Ξ£(Fi x distancei) Each Individual Work Force component in direction of displacement. Displacement component in direction of force.
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Work equals change in Kinetic Energy
Work and Energy πΉβπ₯ cos π = 1 2 π π£ π 2 β 1 2 π π£ π 2 ππππ=βπΎπΈ Work equals change in Kinetic Energy
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Work equals change in energy
Work and Energy Fx cosΟ΄ = Β½ mv2 - Β½ mvo2 Work = ΞEnergy Work equals change in energy
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Examples of Work and Energy
Example 6.7 β Falling rock Use 2nd law Use work Car going down ramp Example Roller coaster Couldnβt do easily by 2nd law! Vertical circle example (use work) Note how you βmix upβ dimensions!
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More Examples of Work and Energy
Example 6.6 β Work to increase car speed Example 6.5 β Work to stop car Problem Air resistance on baseball
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October Potter County hiking / camping
Outta here October Potter County hiking / camping
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