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Physics 218 Exam 2 Fall  2010,  Sections  521-­524  

 

Do  not  fill  out  the  information  below  until  instructed  to  do  so!  

 

Name   :____Solutions  in  RED_____  Signature   :______________________________  

Student  ID   :______________________________  

E-­mail   :______________________________  

Section  #   :______________________________    

Rules  of  the  exam:  

1. You  have  the  full  class  period  to  complete  the  exam.  2. When  calculating  numerical  values,  be  sure  to  keep  track  of  units.    3. You  may  use  this  exam  or  come  up  front  for  scratch  paper.    4. Be  sure  to  put  a  box  around  your  final  answers  and  clearly  indicate  your  work  to  your  grader.  5. Clearly  erase  any  unwanted  marks.  No  credit  will  be  given  if  we  can’t  figure  out  which  answer  you  are  

choosing,  or  which  answer  you  want  us  to  consider.  6. Partial  credit  can  be  given  only  if  your  work  is  clearly  explained  and  labeled.  7.  All  work  must  be  shown  to  get  credit  for  the  answer  marked.  If  the  answer  marked  does  not  

obviously  follow  from  the  shown  work,  even  if  the  answer  is  correct,  you  will  not  get  credit  for  the  answer.  

Put  your  initials  here  after  reading  the  above  instructions:__________  

 

 

 

 

 

 

 

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Table  to  be  filled  by  the  graders  

Part   Score  Part  1  (30)    Part  2  (35)    Part  3  (35)    Part  4  (25)  Bonus  points  

 

Bonus  (5)    Exam  Total    

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Part  1: (25p)  Basic  Concepts  Problem  1.1: (3p)  Fill  the  following  table  

Quantity   Units   Is  it  a  vector  ?,  or  a  scalar  ?  

Force   Newton   Vector  

Work   Joules   scalar  

Power   Watt   scalar  

Energy   Joules   scalar  

 Question  1.1.1: Two  masses  are  attached  by  a  spring  and  moving  with  different  velocities  to   the   right   in   a   surface   with   friction.   One   of   the   masses   has   a   string   attached   to  another  mass  via  a  pulley.    

Question  1.1.2:  (5p)  Below  draw  the  free-­‐body   diagrams   of   all   three  masses   (A,B  and  C)  giving  the  forces  distinct  names.  

 

 

 

 

 

 

 

 

 

Question  1.1.3: (5p)   Considering   only   the   forces   drawn   in   the   previous   question,  write   in   the   table  below  the  names  of  ONLY  those  forces  in  the  three  free-­‐body  diagrams  that  are  action/reaction  pairs.  

 

   

 

 

 

 

 

Action   Reaction  FS  on  A   FS  on  B  Fr  on  B   Fr  on  C                  

g  

friction  

A   B  

C  

wC  

Fr  on  C  !"#! !"#!

NA  

wA  

Fs  on  A  

f  on  A  !"#! !"#!

NB

 

wB  

Fs  on  B  

Fr  on  B  

f  on  B  !"#!

!"#! !"#! !"#! !"#!

!"#!

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Problem  1.2: A  particle  is  subjected  to  a  single  force,  whose  potential  energy  is  drawn  in  the  table  below  as  a  function  of  position.  The  particle’s  total  energy  is  12  Joules.  

 

Question  1.2.1: (3p)   If   the   particle  is   initially   at   position   40cm,  roughly   what   is   the   range   of  positions   this   particle   can   ever  reach  ?  

 

From  about  35  cm  to  about  42  cm.  

 

Question  1.2.2: (3p)   Quantum  Mechanics   effects   allow   a   small  probability   for   the   particle   to  reach  the  position  at  28cm.   If   this  happens,  what  is  the  new  range  of  positions   the   particle   can   ever  reach  ?    

 

 

From  about  28cm  to  about  11  cm.  

 

Question  1.2.3: (3p)  Based  on   the  graph  estimate   the  magnitude  and  direction  of   the   force  over   the  particle  when  it  is  at  12  cm.  Explain  your  reasoning.  

The   slope   of   the   line   is   about   8   Joules   in   y   for   every   4   cm   in   x.   Therefore   the   slope   is   -­8Joules/4cm  =  -­2  Joules/cm=  -­200  Joules/m.    Because  the  force  is  minus  the  slope  we  get  F=200  Joules/m=  200  Newton  in  the  positive  x  direction.    

 

 

 

Question  1.2.4: (3p)   Is   the   point   at   18   cm   a   point   of   equilibrium?,   if   yes   is   it   stable   or   unstable   ?  EXPLAIN  WHY!!    

It  is  a  stable  point  of  equilibrium  as  small    variations  of  position  will  tend  to  bring  the  point  back  to  equilibrium.  

 

0                                  8                    16                  24                32                  40                    48                56                  64                  72  

56  

48  

40  

32  

24  

16  

8  

 

U  (Joules)  

position  (cm)  

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Part  2: (25p)  Forces  and  Acceleration  Problem  2.1: An   object   of  mass  m   is   connected   to   a   spring   nailed   to   a   board   and   undergoing  uniform  circular  motion  in  the    horizontal  plane.  Question  2.1.1:  (5p)  Draw  the  free-­‐body  diagram  of  the  mass  m.  

 

 

 

 

Question  2.1.2: (15p)  It  is  known  that  the  when  the  radius  of  the  circular  motion  is  R1  the  period  of  revolution  is  T1,  and  when  the  radius  of  the  circular  motion  is  R2  the  period  of  revolution  is  T2.  Find  the  constant  k  of   the  spring  and  its  

natural  length  l0.  

€

F = ma⇒

−k(R1 − l0) = −m 4π 2R1

T12

−k(R2 − l0) = −m 4π 2R2

T22

From the first equation we get,

k = m4π 2R1

T12(R1 − l0)

, which plugging into the second we obtain

m4π 2R1

T12(R1 − l0)

(R2 − l0) = m 4π 2R2

T22 ⇒ R1

T12(R1 − l0)

(R2 − l0) =R2

T22 ⇒ R1T2

2(R2 − l0) = R2T12(R1 − l0)

⇒ R1T22R2 − R2T1

2R1 = l0(R1T22 − R2T1

2)⇒ l0 =R1R2(T2

2 −T12)

(R1T22 − R2T1

2), and k = 4π 2m(R1T2

2 − R2T12)

T12T2

2(R1 − R2)

 

Question  2.1.3: (2p)  Express  the  kinetic  energy  of  the  mass  in  terms  of  the  spring  constant  k,  the  radius  of  revolution  R,  and  the  natural  length  l0.  (No  need  to  substitute  the  values  you  got  from  the  previous  question).    

€

−k(R − l0) = −mv 2

R⇒

mv 2

2= K =

Rk2(R − l0)  

Question  2.1.4: (3p)  Express   the   ratio  of   the  Kinetic  Energy   to   the  spring  potential   energy   in  terms  of  R,  and  the  natural  length  l0.  

 

€

K =Rk2

(R − l0), U = k2

(R − l0)2 ⇒KU

=Rk(R − l0)k(R − l0)2 =

R(R − l0)  

 

m  

 

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Part  3: (25p)  Forces  and  Energy  Conservation  Problem  3.1: Two  objects  of  the  same  mass  m  are  attached  by  a  spring  of  constant  k  and  on  the  frictionless  surface  of  respective  inclined  planes.  The  inclined  planes  are  joined  at  one  end  with  an  aperture  angle  of  α  as  shown  in  the  figure  below.  

Question  3.1.1:  (10p)   Knowing   that   the  two  masses  are  in  equilibrium  at  height  h0.  Use   Newton’s   2nd   law   to   find   the  distance   of   compression   of   the   spring  in   term   of   h0,  m,   g,   k,   and  α .   From   it  then   obtain   the   natural   length   of   the  spring.  

€

X : T - Ncos(α) = 0⇒ N = Tcos(α)

=kΔX

cos(α), with

ΔX being the compression distanceY : Nsin(α) - mg = 0 ⇒ kΔX tan(α) - mg = 0

⇒ ΔX =mg

k tan(α)  

€

The geometry of the problem shows that R0 = h0 tan(α) and ΔX = l0 − 2R0 = l0 − 2h0 tan(α)⇒

l0 = 2htan(α) + ΔX = 2h0 tan(α) +mg

k tan(α)  

Question  3.1.2: (5p)  Find  the  total  energy  of  the  system  of  both  masses  at  the  equilibrium  point  with  the  coordinate  system  indicated  in  the  figure.    

€

E = 2mgh0 +k2ΔX 2 = 2mgh0 +

k2

(mg)2

k 2 tan2(α)  

Question  3.1.3: (10p)   Somebody   kicks   the   system   in   equilibrium   giving   the   two   masses  simultaneously  a  velocity  v0  in  the  downward  direction.  At  some  point  later  the  system  is  at  maximum  compression  and  the  mass  velocities  are  zero.  Using  conservation  of  energy  write  an  equation  from  which  it  is  possible  to  obtain  the  new  height  h  at  which  this  happens.  The  equation  must  depend  only  on  h0,  m,  g,  k,  l0,  v0  and  α .  DO  NOT  SOLVE  THE  EQUATION.  

€

After the kick has been given, the energy of the system shall remain constant.

Ei = 2mgh0 +k2

(mg)2

k 2 tan2(α)+mv0

2 initial energy just after the kick.

E f = 2mgh +k2

(2h tan(α) − l0)2 energy at maximum compression

Equating Ei = E f gives an equation for which the new h can be found,

 

 

g  

X  

Y  

m   m  

α

h0  

!

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Part  4: (25p)  Static  Friction  Force  Question  4.1.1: A  block  of  mass  m    is  atop  the  surface  of  a  cart  at  an  angle  α  with  respect  to  the  horizontal.  There   is   friction  between   the  mass  m   and   the  surface  of   the  car,   in  such  a  way  that  the  block  does  not  slide  with  respect  to  the  surface  of  the  cart  when  it  is  at  rest.    

Question  4.1.2: (5p)   Draw   the   free  body  diagram  of  the  mass,  including  angles.  

 

 

 

 

 

Question  4.1.3: (15p)  If  the  cart  is  accelerating  in  the  positive  X  direction  with  acceleration  ax,  and  the  block  is  still  not  slipping  with  respect  to  the  surface,  find  the  magnitude  of  the  force  that  the  plane  exert  on  the  block  (N),  and  the  actual  magnitude  of  static  friction  force  (fs)  in  terms  of  m,  α ,  and ax.  (Hint:  write  Newton’s  equations  in  terms  of  N  and  fs)  

€

X : - N sin(α) + f s cos(α) = max

Y : Ncos(α) +f s sin(α) - mg = 0

From the second equation we get ⇒ N =mg − f s sin(α)

cos(α) and plugging into the first one

− mg − f s sin(α)( ) tan(α) + f s cos(α) = max ⇒ f s cos(α) +sin(α)tan(α)( ) = m ax + gtan(α)( )

⇒ fs =m ax + gtan(α)( )

cos(α) +sin(α)tan(α)( ) and therefore N = mg

cos(α)− tan(α)

m ax + gtan(α)( )cos(α) +sin(α)tan(α)( )

 

 

Question  4.1.4: (5p)   Now   that   you   know   the   normal   force   N   and   the   friction   force   fs   as   a  function  of  m,  α ,  and ax   .    If  the  value  of  µs  is  known,  explain  in  words  how  would  you  find  the  minimum  acceleration  that  makes  the  block  slip.  

The  block  would  slip  once  the  acceleration  is  such  that  the  needed  friction  force  to  keep  it  in  place   is   larger   than   µsN,   and   so   we   need   to   find   the   acceleration   ax   such   that     makes  fs(ax)/N(ax)  greater  than  the  given  value  of  µs  

 

 

 

g  

X  

m  friction  µs   α

!

N  

mg  

fs    α

α