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 Particular cases: (i) φ = 90 0 → cos φ = 0 → W = 0 (no work) (ii) φ = → cos φ = -1 → W = - Fs ≤ 0 (negative work) James Joule (1818 – 1889)

Kinetic Energy and Work-Energy Theorem

Work and Energy with Varying Force (1D-motion) F x =const → W = F x (x 2 – x 1 ) Particular cases: F x = -k·x (Hooke’s law) →

Work-Energy Theorem for 1D-Motion under Varying Forces X m is kinetic energy Example 6.7: Air-track glider attached to spring Data: m=0.1 kg, v 0 =1.5m/s, k=20 N/m, μ k = 0.47 Spring was unstretched. Find: maximum displacement d Solution:

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

Power Average power Instantaneous power Units: [ P ] = [ W ] / [ T ], 1 Watt = 1 W = 1 J / s 1 horsepower = 1 hp = 550 ft·lb/s = 746 W = kW Related energy unit 1 kilowatt-hour = 1 kWh = (1000 J/s) 3600 s = 3.6·10 6 J = 3.6 MJ Power 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-energy theorem. James Watt ( ), the developer of steam engine.

Exam Example 13: Stopping Distance (problems 6.29, 7.29) x 0 Data: v 0 = 50 mph, m = 1000 kg, μ k = 0.5 Find: (a) kinetic friction force f kx ; (b)work done by friction W for stopping a car; (c)stopping distance d ; (d)stopping time T; (e)friction power P at x=0 and at x=d/2; (f)stopping distance d’ if v 0 ’ = 2v 0. Solution: (a)Vertical equilibrium → F N = mg → friction force f kx = - μ k F N = - μ k mg. (b) Work-energy theorem → W = K f – K 0 = - (1/2)mv 0 2. (c) W = f kx d = - μ k mgd and (b) yield μ k mgd = (1/2)mv 0 2 → d = v 0 2 / (2μ k g). Another solution: second Newton’s law ma x = f kx = - μ k mg → a x = - μ k g and from kinematic Eq. (4) v x 2 =v a x x for v x =0 and x=d we find the same answer d = v 0 2 / (2μ k g). (d) Kinematic Eq. (1) v x = v 0 + a x t yields T = - v 0 /a x = v 0 / μ k g. (e) P = f kx v x → P(x=0) = -μ k mgv 0 and, since v x 2 (x=d/2) = v 0 2 -μ k gd = v 0 2 /2, P(x=d/2) = P(x=0)/2 1/2 = -μ k mgv 0 /2 1/2. (f) According to (c), d depends quadratically on v 0 → d’ = (2v 0 ) 2 /(2μ k g) = 4d

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