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Name _ AP Advanced Physics Energy Written 1. Problems A 1200 kg elevator driven by an electric motor can safely carry a maximum load of 800 kg. What is the power provided by the motor when the elevator ascends with a full load at a speed of 2.3 m/s? (45kW) t:::- ~ :: fV (\'LCO 4-~D(:)(q.«'Xj..~) -\::: 2. -= \4$ PN \ ----~ A small object of mass m slides without friction around a loop-the-Ioop apparatus as shown below. It starts from rest at point A at a height h above the bottom of the loop. (a) What is the minimum value of h (in terms of R) such that the object moves around the loop without falling off at the top (point B)? (5RI2) eli o-.t \ '2. ,1-1" "I '"~'I'D::,". " hI' 1.t" ~:q;;'L~~~ "'" __ u 1. t.?l 'v\ :::-""~ -\- ~::: L y- \J : " 'Zoo, - (b) If h = 3R, compute the following quantities horizontal diameter: (i) Velocity (2(grf5) 2- ~ ~~;\ L."(" + ~ "l. ~ ::: 2.'("'r for the object at point C, which is at the end of a 0- ~"":. 7--j ""c.. '2('J:R '12..-;: (ii) -jr) radial ac~e.J:ration ~4g) and 0. yo -= r 1- l1. JY"') -:= -6] ("" (ii i) tangential ~~ "(V\~ acceleration (g) "1'C1 -= ~ ~o.."'<:. -= \N '(Y""\ 0,...... rio..;,. :C\\ 3. A ball of mass 0.5 kg is tied to a string of lengt~5-m~he other end of the string is tied to a rigid support. The ball is held so that the string makes an angle of 60° with the vertical, with the string pulled taut. The ball is then released from rest. 2•.•..• (a) What is the speed of the ball at the lowes,t point o{ its motion? Include a diagram in your ~ =*~" T solution. .s¥-~rJ 2~~~~\=~ >v/.o. h (4.43m/s) \ -= ::;.<- rr." V ~ f2-i-'1-",,~i N\ __ -= J1. (el.i') (') -= ----.] t LJ • 4-3 ? (b) What is the tension in the string when the ball is at the lowest point of its motion? Include an FBD in your solution. (9.8N) m"2.. ,-VV -:: - , B R ~ _") '2'(" -/ R J :::') ,,'l- -\- t1 (oJ: s?-v.c\ .r.. :z. 4. A certain spring is faund NOT to obey Hooke's law, but rather exerts a restoring force F(x) = - 40 x - 9 x' if it is stretched orcompressed a distance x. The units of the numerical factors are such that if x is in meters, then F will be in newtons. (a) Calculate the potential energy function U(x) for this spring. Let u=o when x=O. (20x'+3x3) r: _ - d \A. t...4O'i - q)( '2.) d;<. == - 1\.»- C di l~\,,*,'\"') \ V"= LO'/..1.. -+ ~)(~] (\<\Ic<.t~) c\1A. iy -4 o~+ ~Yo1- -=- -- 6.y., J c\.)I. -= -~ 0)( _ ~ )(.'Z.) _ w~ d. v.. _L_. ------' \c-::: - 'l. - "5 'J. \A An object of mass~ IS attached to this spring, pulled a distance 1.2 m to the right an a frictionless horizontal surface and released. What is the velocity of the object when it is 0.5 m to the right of the Q-\1;<=oequilibrium position?(4.37 m/s) \.1fl'. tA..-='LaC 1.2.) --I 3l,o'2..)2 -= ~~.C\~l.f-S (<;"<'?A(~ \)w.-\o.:d-".l (1;) c.\v... 1. \A-=00~.S) -, ;-;(.sy _ c;.?:>153' :: 2- ~ .u>O(1":S ~J \ v-= 4.'~1 5. The patential energy function of a particle is Xv. 1, ~+-__ -') 0 Cl S 10 (-ICO0) ~ ('300'))('2;-(-\2))(.:',. 1-"" U(x)=-k:c+bx+c 2-50 2 where k = 300 N/m, b = -12.0 N/m' and c = -1000 J. ULlI.)::\SO( f ~;. -\2)( -IOI::>D (a) Plot the function fram x = -5.00 m to x =+ 15.0 m. attach a copy of the graph to this packet. 1-z.'5~ (b) Assume the particle has a mass of 50 kg and is released from rest x = -5.0 m. Describe the ~cJ:> subsequent motion of the particle. Where is the particle movingfastest? Slowest? Is it ever at -\000 IS ;\1':>0 Detwt-tY'\ re;t? -S-D l'l\Q!(2-) II 0 - S ~. cUcx-eo..';,.{s. ct\{)~: ro uecP'''St:S ~(.-e \-00 \-o~l~ \'<\C(-eW~ \-t f'C'~. C,'N\~'f\ \A 't' l;o ¥-€" =- ') ~I'""\ =)I.A..,'bc¥-t't ~W-t ~ 'i?-\ S II f 'I'nt<~W ~\C\\~ (c) Assume the particle is released from rest at x = +5.0m. Describe the subsequent motion of the (1A.li particle. ~~et') (d) If the particle is released at x = -5.0 m with a speed of 10 mis, how fast will it be going at x = 0.8 I,.L. m? (17.5 m/s) ~\ ~()6 \ l.J'\1..oU<- -'5 rY" \ -= v..CJ + '(.. -=- '-12'50 It,1S0:) -= 1. rf'V lP1C;O 7.. -t :::1-5N'l.-\- l _---J vl':SOfq\O.\y'\ ~'S ()L ') -;::-..L (COO"" V"2. + 7- 9ty- _ ~.\Lj -;:-:.JL. v ( ~( 0)2.. -+ '2. 50) 1 1'5 D (.. -=:::. q):L _ ~ I SO J 11...'• '/)?> -I \,; 'J -1000 I Ih' V-::"\J'bOv.T ~S'" =~ COt) (j;).. A tow truck is pulling a car of mass M=2000kg attached vdth a cable as shown. There is also frictional force on tbe Cill. At position x=O the car bas velocity Vo = 5m/s. ~~@ z: __ %&0 a. In the space below forces acting on it. draw the free body ~.~ diagram for the car, showing all ~_L\ ~rv-~ b. What is the kinetic energy KE of the car at x=O? I 2..'5pOOT IKE= :~'J••.. 1 If the total frictional force is a constant Ff ::: SON. what is the work Wf done by friction on the car when it moves from x ::: 0 to x =150 m? c. I_ l,500:J d. "1-.', The low truck pulls the cable so that the tension T in the cable ~J 300 lie .s changes. as shown in the graph. 200 What is the work WT done on the ~ car by the tension force when a = the car moves ,,=150 m. from x=O to ~ 100 o W-==t=.X \ 'SQOQ e. + o 15'00 50 100 150 200 position x In m IWT=22 1500:r N." '" f:::,'.; € 2./L,SDO -l,'SOO -= ~ I '{\l -2S',UQoKE= ~.c:: /'5,000 I .)..;;J:.SLj What is the kinetic energy KE of the car at x = 150m? i-fO,Ot)OT +'2.$',000 f. What is t'he' power PT supplied to the car by the tension force initial point x = 07 at the ------- - -- ---_.- A net force acting on a glider on an air track varies with its position x along the track as shown in the graph. The glider, which has mass = O.Bkg. moves without friction. Force in N 6 o 2 x in m -6 (b) If the velocity of the glider is zero at x = O. find the kinetic energy K and tile velocity v of the glider at the position x = 1m. \f'J -:: tX-te :::'3 -= ~t:'- 0 2.( 31')-= l~~ V'2.j 2 r..o -= (-.-t) V '"l~N{ \/'1-= ~ I _K=_;,_j I ~' ' G T) I---!O.-l =)\j::J{ t]. .-,t-t s (i. ht1ti~) v= (c) Suppose now that the initial velocity of the glider at x:::: 0 is 0.5 mls. The same force shown in the graph above the acts on the glider. What are the kinetic energ.v K and the velocity v of the glider at the position x = 1m. 0'\ [) ) A mass m is lied to a string of length L. The mass is released {rom rest at a height h as shown in the figure. Give your answers below as simple expressions in terms of the symbols defined and the acceleration of gravity g. P e_--"! I '" 1(;\ -~O B 3. Find the speed VB of the mass at the bottom marked B on the figure. ~~ of its trajectory, ~ ~ rf-v'- = 2..3 h . \J 2. , Y::.J~Cjn' b. When the mass is at the point B. -the string hits a is a distance d above B. The mass continues around d to the point P where the string is horizonlaI. Find the mass at point P. ¥-c 'f. E t' i. rf rt-S '<\) ~ 2 ~ II . v;} ? 1 ! 2" -='..J CJ\.tJ ! -= Ii gu re. D--y- :: Cl; V vp ~d = J 2(3""' -<j d) 1 1Jj (Y, -,,) "2- P 'l-\ - l ~#- ...lr 0..\ -'<l c\ d z~~~ ;''r#.~... 'd.' \Vh;{ I c;\) ,_V 7- _ '2:..:.J0->-(~_- ...•• o...0=-C4 A -:.JIlI.1-{().('V\, .1.. -\ rf'jq ~ c. at is the acceleration vector ap of the mass al the point P? Give your answ~r in terms of the unit vectors I and j shown in the vY'~~c:\(" o J CI-"2. SI; ~T\ :i Vp"L ~ '- - t> <.----.- I, 'v'I Y'j\ Vrr'- ~d\ '?:.( ~'" V -:::.J 1..l5"" - ~ d) If' '2. =: T U ':lp (?" -;:: t fixed rod which a circle of radius the speed vp of ,. is [he tension TI, l.'3CY--d)1--+;'I .. ----:--:c:..=J' -d Tt'zt" y '' J T in the sIring when the mass is at point ;s. ~~~ __ 0 -:1 'l-I T-:;'(Y\~q(h-dD T= r~j:' P? -:UY'O}_ v-..-o.') - d ., /j~ Q \]JWr-<' V-= 0~-dV2 J2~Ch-d) '2'jC'"'-<0- C;-O\\l e -0>( d VJ -\1> ------- f1, -I ) . A skier starts from rest at th(>top ora ski jump as sho •••. 'll. (Neglect Friclion.} A h B= 10 ~r-- ::::r:e = C B 15' ' -----r--- he: ::'5m Ground Level If the speed of the skier at point C is 20 mls. what is the speed 3. u:ihe track. (B) Tott"'" 'f.c .l-if '2. ('2.0)7- -::¥-\b 1- 1- ~t +?"q (1'5) 2-00 -+ 1'1' - -:: ~ 'H' -= VB at the bottom v,g i~ VEo~ ~ + ';0a(lo) '5'l. J b. \\1':2t is the starting height h,\? rr< ~'" "" := ''t\p.. -=- 'rf\ l'2..t> ~ 2.P 0 y- .HY\ Ij 15 -'r \'11 I [h =3'5.4YY") "I,r A :5 s.. '1 i},? c. At point C the track ends v."'ith a slope of 15C• How high v.ill tl:e skier rise abo'!:€' troe ground after leaving the track? mfA-X \If ~%~ V'1r,.= 0 V'j:-:: 1-: v ( 'l-+ '2"3 t:>)l. ( c; i\~) ::::bY.. = \.~..,"" -I- ISm '2..(0..«\ d. tt the mass I hm •• = I~.4/V1 I 110, 'j rJ1 of the skier is 75 Kg. and the radius of curvature of :.he traCk at B i~ R=30 m. what is force F orthe track on the skier at point B? IF= \q lKN I I 'i -f'>.; ---- .-- _--- •. I\.i