CHAPTER 6 -- ENERGY. Approach #2: Using the component of mg along the line of d:

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1 Slutins--Ch. 6 (Energy) CHAPTER 6 -- ENERGY 6.) The f.b.d. shwn t the right has been prvided t identify all the frces acting n the bdy as it mves up the incline. a.) T determine the wrk dne by gravity as the bdy mves up the incline, there are tw appraches. Fr yur cnvenience, the frce, velcity, and displacement are pictured belw and als t the right. = 5 f k N d =35 m v = m/s F mg Apprach #: Using the definitin f wrk and the angle between mg and d as φ : W grav = F. grav d = (mg) (d) cs φ = (mg) (d) cs (5 + 9 ) = (3 kg)(9.8 m/s )(35 m) (-.43) = jules. mg d =35 m = 5 Apprach #: Using the cmpnent f mg alng the line f d: W grav = F. grav d = +(F mg parallel t "d" ) (d) = -(mg sin 5 ) (d) = -[(3 kg)(9.8 m/s )(.43)] (35 m) = jules. mg d =35 m = 5 mg sin (cmp. f "mg" alng line f "d") Nte: In this case, the frce is OPPOSITE the directin f the displacement which means the wrk must be negative. The negative sign in this case must be inserted manually. An alternative wuld be t ntice that the angle φ between d and F's cmpnent alng d's line is 8 and determine the wrk quantity using: 499

2 W grav = F. grav d = (F mg parallel t "d" ) (d) cs φ = (mg sin 5 ) (d) cs 8. In ding s, the csine functin will give yu the - autmatically. b.) The frictinal frce is equal t µ k N. T determine N, we need t use an f.b.d. and N.S.L. in the nrmal directin. The f.b.d. is shwn t the right. N.S.L. yields: N F F N : N - mg cs θ = (as a N = ) N = mg cs θ f k = µ k N = µ k (mg cs θ) = (.3) (3 kg)(9.8 m/s ) cs 5 = 7.99 nts. f k mg mg cs mg sin Frictin is always ppsite the directin f mtin. The wrk frictin des will be: W f = f. k d = (f k ) (d) cs 8 = - f k d = -(7.99 nts)(35 m) = jules. c.) The angle between d and N is 9. The csine f 9 is zer. That means that the net wrk dne by the nrmal frce will be zer... ALWAYS! d.) Kinetic energy is defined as (/)mv. Using that expressin we get: = (/)mv =.5 (3 kg) ( m/s) = 6 jules. 5

3 Slutins--Ch. 6 (Energy) e.) The wrk/energy therem states: W net = KE. Fr this case, that means: W net = KE. W f + W F + W mg = KE -. (f k ) (d) cs 8 + F (d) cs + (mg) (d) cs φ = (/)mv - (/)mv. Plugging in the numbers, we get: ( J) + F(35 m) + ( J) = (/)(3 kg)(7 m/s) - (/)(3 kg)( m/s) F =.34 nts. psitining f frce F when mass is at an arbitrary angle 6.) The situatin is shwn in the sketch t the right. We need t derive a F general algebraic expressin fr the frce F acting n the blck, given the fact that that frce is always riented at an angle f m relative t the directin f mtin. Cnsider the f.b.d. (shwn belw) fr the frces acting n the bdy when lcated at an arbitrary angle. As v is small and cnstant, bth the centripetal acceleratin (i.e., v /R) and translatinal acceleratin (i.e., dv/dt) are zer r apprximately zer. Therefre, N.S.L. and the f.b.d. yields: F x : N - mg sin θ + F sin = (as a N = ) N = (mg sin θ) - F sin. N line f mtin F frictinal dme F sin F cs Knwing the nrmal frce, the frictinal frce fllws as: f k = µ k N = µ k (mg sin θ - F sin ). N k mg sin mg cs mg 5

4 F tang : - µ k N - mg cs θ + F cs = - µ k (mg sin θ - F sin ) - mg cs θ + F cs = F = [ µ k mg sin θ + mg cs θ] / [ µ k sin + cs ]. Rewriting this with the cnstants in frnt f the variable expressin, we get: mg F = cs +µ k sin [ µ k sin θ+csθ]. Knwing the frce in general terms, we can use F. dr t determine the wrk the frce des as the bdy rises frm θ = t θ = 6. Nting that: --The angle φ between the line f F and the line f dr is always ; --The magnitude f a differential displacement dr alng an arc equals R dθ, where R is the radius f the arc and dθ is the differential angle thrugh which the mtin ccurs; --And µ k =., m =.5 kg, and R =.3 meters, we can write: W = F dr = F dr cs φ mg 6 = Rd k + [ µ sin θ+ csθ][ θ] cs cs µ k sin θ= mgr cs = d d k + k 6 [ ] + 6 [ ] cs sin µ sin θ θ = csθ θ µ θ θ= mgr cs 6 6 = k( cs ) (sin ) cs + k sin [ [ µ θ ] + [ θ θ= ] θ= µ ] (. 5 kg)( 9. 8 m / s )(. 3) cs = (. ) cs6 cs (sin 6 sin ) cs + (. )sin [ ( [ ])] + =. 86 jules. [ [ ]] 5

5 Slutins--Ch. 6 (Energy) 6.3) All the energy is stred in spring ptential energy. Using the ptential energy functin fr a spring we get: U sp = (/)kx =.5( nt/m)(. m) =.4 jules. 6.4) The relatinship between the field's ptential energy functin (assumed knwn) and its assciated frce functin is: F( xyz,, ) = Uxyz (,, ) ( U) ( U) ( U) = i + j + k x y z k x e = x k x e k x e i + j + k y z ky ky ky k = x e ky k x ke ky i+ ( ) j 6.5) Lking at the functin, the frce will equal zer when y = and when ln x = 3 (i.e., when x =.86). Using this and ur frce/ptential energy relatinship, we get: U(x, y) - U(x=.86, y=) = - F. dr. Observing that U(x =.86, y = ) = and that dr = dxi + dyj + dzk, and realizing that if this was a test questin, 9% f the pints wuld be wrapped up in the first tw lines f what fllws (i.e., the layut), we can write this as: xy, Uxy (, ) = [( kln( x) 3) i ( ky ) j] dxi+ dyj+ dzk x=. 86, y= [ ] xy, = [( kln( x) 3) dx ( ky ) dy] x=. 86, y= x y = [ k ln( x) 3] dx + [ ky ] x=. 86 dy y= x x y = [ ] + [ 3 ] + [ ] k ln( x) dx dx k y dy. x=. 86 x=. 86 y= 53

6 Setting the k terms equal t ne and nting that (lnx)dx = (x)lnx - x, we can evaluate these integrals as: x x y Uxy (, ) = [ xlnx x] x x. [ ] + 3 x. 3 = 86 = 86 3 y y= y ( ) = [ ( xln x x) (. ln.. )] [([ x] ) ([ (. )])] y = xln x+ 4x Nte: A perfectly legitimate fllw-up questin t this prblem might be: Hw much wrk des the frce field d as the bdy ges frm x =, y = t x = 4, y = -. The answer is: W cns.frc = - U = -[ U pt - U pt ] = -[[-4ln4 + 4(4) + (-) 3 /3 -.9)] - [-ln() + 4() + () 3 /3 -.9)]] = jules. 6.6) a.) We culd use the wrk/energy therem n this prblem, but the mdified cnservatin f energy equatin is s much easier t use that we will use it here. Nting that the tensin in the line is always perpendicular t the mtin (i.e., the wrk dne due t tensin is zer), we can write: + U + W extra = KE + U (/)mv + mgy + Tdcs 9 = (/)mv + mgy () + m(9.8 m/s )( m) + () = (/)mv + m(9.8 m/s )(5 m). Being careful nt t cnfuse mass terms dented by m and the units f length (meters, abbreviated m), we can cancel the mass terms and get: v =.7 m/s. b.) At the bttm f the arc, Tarzan's velcity can again be fund using the mdified cnservatin f energy expressin (we need that velcity 54

7 Slutins--Ch. 6 (Energy) because he is mving under the influence f a center-seeking frce--a velcity driven functin--at that pint). Using the apprach: + U + W extra = KE bt + U bt (/)mv + mgy + () = (/)mv bt + mgy () + m(9.8 m/s )( m) + () = (/)mv bt + m(9.8 m/s )( m). Canceling the mass terms yields: v bt = 4 m/s. There is a tensin frce upward and gravity dwnward when Tarzan is at the bttm f the arc (yu might want t draw an f.b.d. t be cmplete). N.S.L. implies: F c : T - mg = ma c = m(v /R) (as a c is a centripetal acc.) T = mg + mv /R = (8 kg)(9.8 m/s ) + (8 kg)(4 m/s) /(5 m) = 89.3 nts. c.) At the mlehill, Tarzan's velcity is.7 m/s. An f.b.d. fr that situatin is shwn belw-left. The nly thing that is really tricky abut the prblem is determining the angle θ. The diagram belw-right will d that. T cs = /5 mg cs mg sin mg 5 m m 3 m 5 m 5 m grund m With θ, N.S.L. yields: 55

8 F c : T - mg cs θ = ma c = m(v /R). As a c is a centripetal acceleratin: T = mg cs θ + mv /R = (8 kg)(9.8 m/s )(/5) +(8 kg)(.7 m/s) /(5 m) = nts. 6.7) Bth gravity and frictin d wrk as the bdy slides dwn the incline. Using the mdified cnservatin f energy equatin, we get: + U + W extra = KE + U (/)mv + mgy + (-f k d) = (/)mv + mgy () + mgr - (f k d) = () + () f k = mgr/d. In this case, d is the ttal distance ver which the frictinal frce acts. That is, the 8 meters alng the hrizntal surface AND the quarter circumference dwn the curved incline (that will equal (/4)( R)). That ttal distance is: d = R = ( m) =.4 m. Plugging this int ur expressin, we get: f k = mgr/d = ( kg)(9.8 m/s )( m)/(.4) =.3 nts. 6.8) The fact that the angle is 85 makes n difference, assuming the velcity is great enugh t allw the dart t make it t the mnkey. What is imprtant is that the dart has enugh energy in the beginning t effect a pierce at the end. Using cnservatin f energy: 56

9 Slutins--Ch. 6 (Energy) + SU + SW ext = KE + SU (/)mv + () + () = (/)mv + mgh. Dividing ut the masses and multiplying by yields: v = [v + gh ] / = [(4 m/s) + (9.8 m/s )(35 m)] / = 6.5 m/s. 6.9) Assuming an average frictinal frce f 7 newtns: a.) Let the ptential energy equals zer level be the grund. That means that belw-grund level, the h in mgh will be negative. Using cnservatin f energy: + U = KE + U (/)mv + () + (-fk d) = (/)mv C + mghc. r.5(8 kg)(38 m/s) - (7 nts)(3 m) =.5(8 kg)v C + (8 kg)(9.8 m/s )(-5 m) v C = 4.64 m/s. b.) The nly thing that is tricky abut this is finding the ttal distance d traveled (we need that t determine the amunt f wrk frictin des). Nting that the distance traveled while mving thrugh the lp is R, cnservatin f energy yields: + U = KE + U (/)mv + () + (-fk d tt ) = () + mgh ramp (/)mv + - fk ( ()+d) = + mgd sin θ.5(8 kg)(38 m/s) - [(7 nts)(95.7 m) + 7d] = (8)(9.8 m/s )d sin 3 d = 46 m. c.) Alng with the cnservatin f energy, this prblem requires N.S.L. We knw that at the tp f the arc, the vertical frces are centripetal at the 57

10 velcity required AND the nrmal frce ges t zer. Frm the cnservatin f energy we get: + U = KE + U (/)mv + () + (-fk d tp ) = (/)mv tp + mghtp (/)mv + - fk [7+6+4+( R)/] = (/)mv tp + mg(r).5(8 kg)v + - (7 nts)(3.8 m) =.5(8 kg)vtp + (8)(9.8 m/s )[( m)] v = vtp Using N.S.L., we get: -N - mg = -ma c = -m(v tp /R). When the velcity is crrect, N ges t zer and: v tp = gr. Substituting that int ur expressin, we get: v = vtp + 79 = gr + 79 = (9.8 m/s )( m) + 79 v = 3.4 m/s. 6.) a.) Assuming the spring is depressed a distance x, let's define the gravitatinal ptential energy equals zer level t be at that pint (i.e., where the spring is depressed a distance x). As such, the crate travels a distance d + x, where d is the crate's initial distance frm the bumper. The mdified cnservatin f energy therem implies: + U = KE + U () + [ U grav, + U sp, ] = () + [ U grav, + U sp, ] () + [mg(d+x) sin θ + () ] - f k (d+x) = () + [ () +.5kx ] mg(d+x) sin θ - f k (d+x) =.5kx (6 kg)(9.8 m/s )(3 + x) sin 55 - ( nt)(3 + x) =.5(,)x. 58

11 Slutins--Ch. 6 (Energy) Rearranging:,x x - 45 =. Using the Quadratic Frmula, we get x =.36 meters. b.) In this sectin, we really are nt interested in x--we want t knw hw much energy the blck has just befre hitting the spring, and hw much energy the blck lses by the time it leaves the spring. In ther wrds, this is really a brand new prblem. As such, let's redefine the zer gravitatinal ptential energy level t be at the pint when the blck is just abut t cme int cntact with the spring. If that be the case, the cnservatin f energy spring and bumper crate's initial psitin d = 3 meters m crate's psitin just befre and after cllisin with bumper (ie. Pint 3) allws us t determine the energy f the blck just befre striking the spring at Pint 3: h = d(sin ) + U = KE 3 + U 3 () + mg dsin θ - f k d = KE 3 + () KE 3 = mgd sin θ - f k d = (6 kg)(9.8 m/s )(3 sin 55 ) - ( nt)(3 m) = jules. If 3/4 f the kinetic energy is lst, then /4 is left. That means the blck has kinetic energy (/4)(44.98 j) = 86. jules as it starts back up the incline. If we let L be the distance the blck travels up the incline t rest, and if we remember that the U = level is at the spring's end, we have: KE 4 + U 4 = KE 5 + U 5 (86. j) + () - ( nt)l = () + (6 kg)(9.8 m/s )(L sin 55 ) L =.49 meters. 59

12 6.) Using the ptential energy functin prvided, cnservatin f energy implies: + U = KE + U (/)m s v + [-Gme m s /(r e + d )] + () = (/)m s v + [-Gme m s /(r e + d )]. Nticing that the m s 's cancel ut, we can write:.5(5 m/s) + [-(6.67x - m 3 /kg. s )(5.98x 4 kg)/(6.37x 6 m+.x 5 m)] =.5v + [-(6.67x - m 3 /kg. s )(5.98x 4 kg)/(6.37x 6 m +.9x 5 m)]. SOLVING yields v = 68 m/s. 6.) The trick t this prblem is in recgnizing the fact that at the tp f its arc, the mass is executing centripetal acceleratin where tensin is acting as ne f the centripetal frces in the system. Using N.S.L. and remembering that the radius f the mass's mtin is L/3, we get: T v tp mg mass at tp f new arc F c : -T - mg = -ma c = -m(v tp /R) = -m(v tp /(L/3)) T = (3m/L)(v tp ) - mg. T slve this expressin fr T, we need the velcity f the mass at the tp f its flight. Enter the mdified cnservatin f energy apprach--an apprach designed specifically t determine velcities when cnservative frce fields are ding wrk n bdies. If we take the ptential energy equal t ZERO pint t be at the bttm f the arc, the infrmatin given in the sketch n the next page tells it all. 5

13 Slutins--Ch. 6 (Energy) INITIAL POSITION AT TOP OF SWING KE = U = mgl attachment W = -.[pre-cllisin energy] xtra = -.[mgl] L/3 v tp KE =.5mv tp U = mg(l/3) peg L/3 peg L/3 U = at arc's bttm Using ur infrmatin, we get: + U = KE + U () + mgl + [-(.)(mgl)] = (/)mv tp + mg(l/3) v tp = [.46gL] /. Plugging this int ur tensin expressin yields: T = (3m/L)(v tp ) - mg = (3m/L)([.46gL] / ) - mg =.38mg - mg =.38mg. 5

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