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Work and Kinetic Energy

Work and Kinetic Energy. Work done by a constant force Work is a scalar quantity. No motion (s=0) → no work (W=0). Units: [ W ] = newton ·meter = N·m = J = joule (SI) [ W ] = dyne·cantimeter = dyn·cm = erg (CGS) 1 J = 1 N · 1 m = 10 3 g 100 (cm/s 2 ) 100 cm = 10 7 erg.

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Work and Kinetic Energy

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  1. Work and Kinetic Energy Work done by a constant force Work is a scalar quantity. No motion (s=0) → no work (W=0) Units: [ W ] = newton·meter = N·m = J = joule (SI) [ W ] = dyne·cantimeter = dyn·cm = erg (CGS) 1 J = 1 N · 1 m = 103 g 100 (cm/s2) 100 cm = 107 erg James Joule (1818 – 1889) Particular cases: (i) φ = 900 → cos φ = 0 → W = 0 (no work) (ii) φ = 1800→ cos φ = -1 → W = - Fs ≤ 0 (negative work)

  2. Kinetic Energy and Work-Energy Theorem

  3. Work and Energy with Varying Force (1D-motion) Particular cases: Fx =const → W = Fx (x2 – x1) Fx = -k·x (Hooke’s law) →

  4. Work-Energy Theorem for 1D-Motion under Varying Forces m X is kinetic energy Example 6.7: Air-track glider attached to spring Data: m=0.1 kg, v0=1.5m/s, k=20 N/m, μk= 0.47 Spring was unstretched. Find: maximum displacement d Solution:

  5. Work-Energy Theorem for 3D-Motion along a Curve Line integral

  6. Power Power is the time rate at which work is done, or the rate at which the energy is changing. These rates are the same due to work-energy theorem. Average power James Watt (1736-1819), the developer of steam engine. Instantaneous power Units:[ P ] = [ W ] / [ T ] , 1 Watt = 1 W = 1 J / s 1 horsepower = 1 hp = 550 ft·lb/s = 746 W = 0.746 kW Related energy unit 1 kilowatt-hour = 1 kWh = (1000 J/s) 3600 s = 3.6·106 J = 3.6 MJ Power is a scalar quantity.

  7. Exam Example 13: Stopping Distance (problems 6.29, 7.29) Data: v0 = 50 mph, m = 1000 kg, μk = 0.5 • Find: (a) kinetic friction force fkx ; • work done by friction W for stopping a car; • stopping distance d ; • stopping time T; • friction power P at x=0 and at x=d/2; • stopping distance d’ if v0’ = 2v0 . 0 x Solution: • Vertical equilibrium → FN = mg → friction force fkx = - μk FN = - μk mg . • Work-energy theorem → W = Kf – K0 = - (1/2)mv02 . • (c) W = fkxd = - μkmgd and (b) yield μkmgd = (1/2)mv02 → d = v02 / (2μkg) . • Another solution: second Newton’s law max= fkx= - μkmg → ax = - μkg • and from kinematic Eq. (4) vx2=v02+2axx for vx=0 and x=d we find • the same answer d = v02 / (2μkg) . • (d) Kinematic Eq. (1) vx = v0 + axt yields T = - v0 /ax = v0 / μkg . • P = fkx vx → P(x=0) = -μk mgv0 and, • since vx2(x=d/2) = v02-μkgd = v02 /2 , P(x=d/2) = P(x=0)/21/2 = -μkmgv0 /21/2 . • (f) According to (c), d depends quadratically on v0 → d’ = (2v0)2/(2μkg) = 4d

  8. Exam Example 14: Swing(example 6.8) • Find the work done by each force if • (a) F supports quasi-equilibrium or • F = const , • as well as the final kinetic energy K. • Solution: Data: m, R, θ WT =0 always since (a) Σ Fx = 0 → F = T sinθ , Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .

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