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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. - PowerPoint PPT Presentation

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• Work and Kinetic EnergyWork done by a constant forceWork is a scalar quantity.No motion (s=0) no work (W=0) Units: [ W ] = newtonmeter = Nm = J = joule (SI) [ W ] = dynecantimeter = dyncm = erg (CGS) 1 J = 1 N 1 m = 103 g 100 (cm/s2) 100 cm = 107 ergParticular cases: (i) = 900 cos = 0 W = 0 (no work) (ii) = 1800 cos = -1 W = - Fs 0 (negative work)James Joule (1818 1889)

• Kinetic Energy and Work-Energy Theorem

• Work and Energy with Varying Force(1D-motion)Fx =const W = Fx (x2 x1)Particular cases:Fx = -kx (Hookes law)

• Work-Energy Theorem for 1D-Motion under Varying ForcesXmis kinetic energyExample 6.7: Air-track glider attached to springData: m=0.1 kg, v0=1.5m/s, k=20 N/m, k = 0.47Spring was unstretched.Find: maximum displacement dSolution:

• Work-Energy Theorem for 3D-Motion along a CurveLine integral

• PowerAverage powerInstantaneous powerUnits: [ P ] = [ W ] / [ T ] , 1 Watt = 1 W = 1 J / s1 horsepower = 1 hp = 550 ftlb/s = 746 W = 0.746 kWRelated energy unit 1 kilowatt-hour = 1 kWh = (1000 J/s) 3600 s = 3.6106 J = 3.6 MJPower is a scalar quantity.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-energytheorem.James Watt(1736-1819), the developerof steam engine.

• Exam Example 13: Stopping Distance (problems 6.29, 7.29)x0Data: v0 = 50 mph, m = 1000 kg, k = 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 .Solution:Vertical equilibrium FN = mg friction force fkx = - k FN = - k mg . Work-energy theorem W = Kf K0 = - (1/2)mv02 .

(c) W = fkxd = - kmgd and (b) yield kmgd = (1/2)mv02 d = v02 / (2kg) . Another solution: second Newtons law max= fkx= - kmg ax = - kg and from kinematic Eq. (4) vx2=v02+2axx for vx=0 and x=d we find the same answer d = v02 / (2kg) .

(d) Kinematic Eq. (1) vx = v0 + axt yields T = - v0 /ax = v0 / kg . P = fkx vx P(x=0) = -k mgv0 and, since vx2(x=d/2) = v02-kgd = v02 /2 , P(x=d/2) = P(x=0)/21/2 = -kmgv0 /21/2 .(f) According to (c), d depends quadratically on v0 d = (2v0)2/(2kg) = 4d

• 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:(a) Fx = 0 F = T sin , Fy = 0 T cos = w = mg , hence, F = w tan ; K = 0 since v=0 .WT =0 always sinceData: m, R,