Problem Solving Steps1. 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, … analysisof the answers and solution.

,...,,, jFavr

,,, zyx

How to measurefriction by meter and clock?

Exam Example 9:

d) Find also the works done on the block by friction and by gravityas well as the total work done on the block if its mass is m = 2 kg (problem 6.68).

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

Exam Example 10: Blocks on the Inclines (problem 5.92)

m1

m2

X

X

α1 α2

1W

1NF

2NF

2W

XW1

XW2

1kf

2kf

Data: m1, m2, μk, α1, α2, vx<0 a

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)

Find: (a) fk1x and fk2x ;(b) T1x and T2x ;(c) acceleration ax .

1T 2T

(a) fk1x= sμkFN1= sμkm1g cosα1 ; fk2x= sμkFN2= sμkm2g cosα2; s = -vx/v (c) T1x=-T2x, Eqs.(1)&(2)→

(b)

21

222111 )sincos()sincos(

mm

sgmsgma kkx

21

12212111121

))cos(cossin(sin)sincos(

mm

sgmmggsamTT k

kxxx

v

Exam Example 11: Hoisting a Scaffold

Ya

T

TT

TT

0

m F

gmW

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

SolutionNewton’s second law: WTam

5

(a) 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)

Data: L, β Find: (a) tension force F;(b) speed v;(c) period T.

Solution:Newton’s second law

j

cj amF

Centripetal force along x: RmvmaF c /sin 2Equilibrium along y: cos/)(cos mgFamgF

cos/sincos/sin

sin,tansin)/()(

2

2

LgLgv

LRRgmFRvb

g

LTLgLvRTc

cos2cos//2/2)(

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

FF

gm

gm ca

ca

Exam Example 13: Stopping Distance (problems 6.29, 7.29)

x

v

0a

Data: v0 = 50 mph, m = 1000 kg, μk = 0.5 Find: (a) kinetic friction force fkx ;(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 v0’ = 2v0 .

NF

gm

kf

Solution:(a)Vertical equilibrium → FN = mg → friction force fkx = - μk FN = - μk mg .(b) Work-energy theorem → W = Kf – K0 = - (1/2)mv0

2 .

(c) W = fkxd = - μkmgd and (b) yield μkmgd = (1/2)mv02 → d = v0

2 / (2μkg) . Another solution: second Newton’s law max= fkx= - μkmg → ax = - μkg and from kinematic Eq. (4) vx

2=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 . (e) P = fkx vx → P(x=0) = -μk mgv0 and, since vx

2(x=d/2) = v02-μkgd = v0

2 /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

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 Fx = 0 → F = T sinθ , Σ Fy = 0 → T cosθ = w = mg , hence, F = w tanθ ; K = 0 since v=0 .

WT =0 always since ldT

Rddsdl

0000

)cos1(sincostancos wRdwRRdwdsFldFWF

0

0 0

)cos1(sin

)sin(

wRdwR

RdwldwWgrav

0 0

sincos)( FRRdFldFWb F

Data: m, R, θ

)cossin( wwFRWWWWK TgravF

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: v

rada

tana

a

(a)To avoid falling off, centripetal acceleration v2/R > g → v2 > gR.Conservation of energy: KB+2mgR=mgh → (1/2)mvB

2=mg(h-2R) . Thus, 2g(h-2R) > gR → h > 5R/2 , that is hmin = 5R/2.

(b) 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.

g

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.

0

ys

yf

y0

y

gm

NF

kf

θ

Solution: work-energy theorem Wnc=ΔK+ΔUgrav+ΔUel

(a)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 (b) (1/2)kys

2 = Uel = ΔUel = Wnc – ΔUgrav = mg (y0+|ys|) (sinθ-μkcosθ) →αys

2 +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 →

k

ks

k

ksf

yyy

yyyy

tan

|)|(2

cossin

cos|)|(2 00

00

Find: (a) kinetic energy and speed at the 1st and 2nd passages of y=0, (b) the lowest position ys and friction energy losses on a way to ys, (c) the highest position yf after rebound. m

Exam Example 17: Proton Bombardment (problem 6.76)

Data: mass m, potential energy U=α/x,initial position x0>0 and velocity v0x<0.

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: (a)Easier way: conservation of energy

(b)Turning point: v(xmin)=0

(c)It is the same since the force is conservative: U(x)=U(x0) v(x)=v(x0)

238U

0 x

mproton

x0

F

xmin

0v

02

xdx

dUFx

22

2

mxdt

xd

xxm

vxUxUm

vxvxUxUmvmv112

)()(2

)()()(2

1

2

1

0

200

200

20

2

020

0min

20

0min 2

2

2

11

xmv

xx

mv

xx