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Problem Solving Steps

Problem Solving Steps. 1. Geometry & drawing: trajectory, vectors , coordinate axes free-body diagram, … 2. Data: a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form , and

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Problem Solving Steps

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  1. Problem Solving Steps 1. Geometry & drawing:trajectory, vectors, coordinate axes free-body diagram, … 2. Data:a table of known and unknown quantities, including “implied data”. 3. Equations ( with reasoning comments ! ), their solution in algebraic form, and the final answers in algebraic form !!! 4. Numerical calculations and answers. 5. Check: dimensional, functional, scale, sign, … analysis of the answers and solution.

  2. Exam Example 9: How to measure friction by meter and clock? d) Find also the works done on the block by friction and by gravity as well as the total work done on the block if its mass is m = 2 kg (problem 6.68).

  3. d)Work done by friction: Wf = -fkL = -μk FN L = -L μk mg cosθmax = -9 J ; work done by gravity:Wg = mgH = 10 J ; total work: W = mv||2 /2 = 2 kg (1m/s)2 /2 = 1 J = Wg + Wf = 10 J – 9 J = 1 J

  4. Exam Example 10: Blocks on the Inclines (problem 5.92) Data: m1, m2, μk, α1, α2, vx<0 m1 Find: (a) fk1x and fk2x ; (b) T1x and T2x ; (c) acceleration ax . X m2 Solution: Newton’s second law for block 1: FN1 = m1g cosα1 , m1ax= T1x+fk1x-m1g sinα1(1) block 2: FN2 = m2g cosα2 , m2ax= T2 x+fk2x+ m2g sinα2(2) α1 α2 X (a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v (c) T1x=-T2x, Eqs.(1)&(2)→ (b)

  5. Exam Example 11: Hoisting a Scaffold Y Data: m = 200 kg Find: (a) a force Fy to keep scaffold in rest; (b) an acceleration ay if Fy = - 400 N; (c) a length of rope in a scaffold that would allow it to go downward by 10 m 0 Solution Newton’s second law: m • Newton’s third law: Fy = - Ty , in rest ay = 0→ F(a=0)= W/5= mg/5 =392 N (b) ay= (5T-mg)/m = 5 (-Fy)/m – g = 0.2 m/s2 (c) L = 5·10 m = 50 m (pulley’s geometry)

  6. Exam Example 12: Data: L, β Find: • tension force F; • speed v; • period T. The conical pendulum(example 5.20) or a bead sliding on a vertical hoop (problem 5.115) Solution: Newton’s second law R Two equations with two unknowns: F,v Centripetal force along x: Equilibrium along y:

  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 .

  9. Exam Example 15: Riding loop-the-loop(problem 7.46) Data: R= 20 m, v0=0, m=100 kg Find: (a) min h such that a car does not fall off at point B, (b) kinetic energies for that hmin at the points B, C, and D, (c) if h = 3.5 R, compute velocity and acceleration at C. D Solution: • To avoid falling off, centripetal acceleration v2/R > g → v2 > gR. Conservation of energy: KB+2mgR=mgh → (1/2)mvB2=mg(h-2R) . Thus, 2g(h-2R) > gR → h > 5R/2 , that is hmin = 5R/2. • Kf+Uf=K0+U0 , K0=0 → KB = mghmin- 2mgR = mgR/2 , KC = mghmin- mgR = 3mgR/2 , KD = mghmin = 5mgR/2. (c) (1/2)mvC2 = KC= mgh – mgR = 2.5 mgR → vC = (5gR)1/2 ; arad = vC2/R = 5g, atan = g since the only downward force is gravity.

  10. Exam Example 16: Spring on the Incline (Fig.7.25, p.231) Data: m = 2 kg, θ = 53.1o, y0 = 4 m, k = 120 N/m, μk = 0.2, v0 =0. Find: (a) kinetic energy and speed at the 1st and 2nd passages of y=0, • the lowest position ys and friction energy losses on a way to ys, (c) the highest position yf after rebound. y m Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel y0 • 1st passage: Wnc= -y0μkmg cosθ since fk=μkFN= =μkmg cosθ, ΔK=K1 , ΔUgrav= - mgy0 sinθ, ΔUel=0 → K1=mgy0(sinθ-μkcosθ), v1=(2K1/m)1/2 =[2gy0(sinθ–μkcosθ)]1/2 2nd passage: Wnc= - (y0+2|ys|) μkmg cosθ, ΔK=K2, ΔUgrav= -mgy0sinθ,ΔUel=0 → K2=mgy0sinθ-(y0+2|ys|) μkmgcosθ, v2=(2K2/m)1/2 yf 0 ys θ (b)(1/2)kys2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) → αys2 +ys –y0 =0, where α=k/[2mg(sinθ-μkcosθ)],→ ys =[-1 - (1+4αy0)1/2]/(2α) Wnc = - (y0+|ys|) mgμkcosθ (c)Kf =0, ΔUel=0, ΔUgrav= -(y0–yf) mg sinθ, Wnc= -(y0+yf+2|ys|) μkmg cosθ →

  11. Exam Example 17: Proton Bombardment (problem 6.76) proton Data: mass m, potential energy U=α/x, initial position x0>0 and velocity v0x<0. 238U m x0 xmin 0 x Find: (a) Speed v(x) at point x. (b) How close to the repulsive uranium nucleus 238U does the proton get? (c) What is the speed of the proton when it is again at initial position x0? • Solution: Proton is repelled by 238U with a force • Newton’s 2nd law, ax=Fx/m, allows one to find trajectory x(t) as a solution • of the second order differential equation: • Easier way: conservation of energy • Turning point: v(xmin)=0 • It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)

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