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. How to measure friction by meter and clock? Exam Example 9: 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: W f = -f k L = -μ k F N L = -L μ k mg cosθ max = -9 J ; work done by gravity: W g = mgH = 10 J ; total work: W = mv || 2 /2 = 2 kg (1m/s) 2 /2 = 1 J = W g + W f = 10 J – 9 J = 1 J

4. Exam Example 10: Blocks on the Inclines (problem 5.92) m 1 m2m2 X X α1α1 α2 α2 Data: m 1, m 2, μ k, α 1, α 2, v x <0 Solution: Newton’s second law for block 1: F N1 = m 1 g cos α 1, m 1 a x = T 1x +f k1x -m 1 g sinα 1 (1) block 2: F N2 = m 2 g cosα 2, m 2 a x = T 2 x +f k2x + m 2 g sinα 2 (2) Find: (a) f k1x and f k2x ; (b) T 1x and T 2x ; (c) acceleration a x. (a) f k1x = sμ k F N1 = s μ k m 1 g cosα 1 ; f k2x = sμ k F N2 = sμ k m 2 g cosα 2 ; s = -v x /v (c) T 1x =-T 2x, Eqs.(1)&(2)→ (b)

5. Exam Example 11: Hoisting a Scaffold Y 0 m Data: m = 200 kg Find: (a) a force F y to keep scaffold in rest; (b) an acceleration a y if F y = - 400 N; (c) a length of rope in a scaffold that would allow it to go downward by 10 m Solution Newton’s second law: (a) Newton’s third law: F y = - T y, in rest a y = 0→ F(a=0)= W/5= mg/5 =392 N (b) a y = (5T-mg)/m = 5 (-F y )/m – g = 0.2 m/s 2 (c) L = 5·10 m = 50 m (pulley’s geometry)

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

7. 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 0 2 +2a 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

8. 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

9. Exam Example 15: Riding loop-the-loop (problem 7.46) Data: R= 20 m, v 0 =0, m=100 kg Find: (a) min h such that a car does not fall off at point B, (b) kinetic energies for that h min at the points B, C, and D, (c) if h = 3.5 R, compute velocity and acceleration at C. D Solution: (a)To avoid falling off, centripetal acceleration v 2 /R > g → v 2 > gR. Conservation of energy: K B +2mgR=mgh → (1/2)mv B 2 =mg(h-2R). Thus, 2g(h-2R) > gR → h > 5R/2, that is h min = 5R/2. (b) K f +U f =K 0 +U 0, K 0 =0 → K B = mgh min - 2mgR = mgR/2, K C = mgh min - mgR = 3mgR/2, K D = mgh min = 5mgR/2. (c) (1/2)mv C 2 = K C = mgh – mgR = 2.5 mgR → v C = (5gR) 1/2 ; a rad = v C 2 /R = 5g, a tan = 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 = 53.1 o, y 0 = 4 m, k = 120 N/m, μ k = 0.2, v 0 =0. 0 ysys yfyf y0y0 y θ Solution: work-energy theorem W nc =ΔK+ΔU grav +ΔU el (a)1 st passage: W nc = -y 0 μ k mg cosθ since f k =μ k F N = =μ k mg cosθ, ΔK=K 1, ΔU grav = - mgy 0 sinθ, ΔU el =0 → K 1 =mgy 0 (sinθ-μ k cosθ), v 1 =(2K 1 /m) 1/2 =[2gy 0 (sinθ–μ k cosθ)] 1/2 2 nd passage: W nc = - (y 0 +2|y s |) μ k mg cosθ, ΔK=K 2, ΔU grav = -mgy 0 sinθ, ΔU el =0 → K 2 =mgy 0 sinθ-(y 0 +2|y s |) μ k mgcosθ, v 2 =(2K 2 /m) 1/2 (b) (1/2)ky s 2 = U el = ΔU el = W nc – ΔU grav = mg (y 0 +|y s |) (sinθ-μ k cosθ) → αy s 2 +y s –y 0 =0, where α=k/[2mg (sinθ-μ k cosθ)], → y s =[-1 - (1+4αy 0 ) 1/2 ]/(2α) W nc = - (y 0 +|y s |) mgμ k cosθ (c) K f =0, ΔU el =0, ΔU grav = -(y 0 –y f ) mg sinθ, W nc = -(y 0 +y f +2|y s |) μ k mg cosθ → Find: (a) kinetic energy and speed at the 1 st and 2 nd passages of y=0, (b)the lowest position y s and friction energy losses on a way to y s, (c) the highest position y f after rebound. m

11. Exam Example 17: Proton Bombardment (problem 6.76) Data: mass m, potential energy U=α/x, initial position x 0 >0 and velocity v 0x <0. Find : (a) Speed v(x) at point x. (b) How close to the repulsive uranium nucleus 238 U does the proton get? (c) What is the speed of the proton when it is again at initial position x 0 ? Solution: Proton is repelled by 238 U with a force Newton’s 2 nd law, a x =F x /m, allows one to find trajectory x(t) as a solution of the second order differential equation: (a)Easier way: conservation of energy (b)Turning point: v(x min )=0 (c)It is the same since the force is conservative: U(x)=U(x 0 ) v(x)=v(x 0 ) 238 U 0 x m proton x0x0 x min