QUIZ 9 (ANSWERS) CH. 7: Problems 9*, 24*, 38* (TRY  #34*,  critical for Ch 12 and Physics 4B; see class notes on  stable and unstable equilibrium)  41*, 42*, 43*, 46* (try #12*), 47* (TRY  65* and 66*) , 50*, 55*, 56*, 63* (see #46*) , 68*, 70*,  75* (SEE  SAMPLE TEST 3) (ANSWERS) 
VIRTUAL LAB 2  (SEE EXAMPLE 7.7, CHAPTER 7)
SIMPLE HARMONIC MOTION LAB 
Sample Exam Cross Index for TEST 2 Sp '12

BELOW ARE THE LINKS TO SAMPLE EXAMS:
For many of these problems use  KEi + Ui  = KEf + Uf + HEAT    or  USE  KEi + Ui  + WF  = KEf + Uf + HEAT, where  U is the potential energy due to the conservative force ( gravitational  and spring forces in this case), KE is the kinetic energy and   WF is the work done by the external force  (done by "Frankie" as we discussed in class. ) 
NOTE:  9, 46 and 63 mix ENERGY CONSERVATION (Ch. 7)and CIRCULAR  DYNAMICS  (Ch. 5), a special class of problems you should study ahead of  Test 2.  
ANOTHER NOTE: #50 mixes ENERGY CONSERVATION (Ch. 7) and PROJECTILE MOTION (CH. 3)
9.* KEi + Ui  = KEf + Uf + HEAT   . 
(a) The angle between the normal force and motion in this case is 90 degrees,  so what is the work by that force? The work by gravity is positive since the motion is from high to low; note the gravitational  work only depends on the vertical  displacement. 
(b) ENERGY CONSERVATION : KEi + Ui  = KEf + Uf + HEAT   where HEAT IS THE NEGATIVE OF THE WORK BY FRICTION. THUS HEAT IS ALWAYS POSITIVE. You are given the heat. 
(c) CIRCULAR  DYNAMICS: Critical point.  HEAT is the negative of the work  by  the friction force, which may  not be constant. The instantaneous friction force is  u*N, where u is the kinetic friction coefficient  and where N is the magnitude of the normal force which may not be constant as in this case. Clearly from earlier problems in Chapter 5, N is a maximum at the bottom and is a minimum at the top.  (See worksheet 2, GRP 3,  problem #1,  THE PROBLEM DEALING WITH BUMPS AND VALLEYS) 
(d) CIRCULAR  DYNAMICS: At bottom, the normal force points up in the pos direction toward the center of the circle and the weight of magnitude mg points down in the neg direction.  Write:
Sum of forces in radial  direction = m*v2/R = pos - neg and solve for N, the  normal force  magnitude,  using the speed v you found in part (b) from  energy conservation. See worksheet 2, GRP 3,  problem #1. Note N > mg.
24. *Use this:   KEi + Ui  = KEf + Uf + HEAT for the initial (i) and final (f) locations.   Study carefully  the classic example (i.e. Many physics book authors  have used it.).  Unlike the previous problem, the friction force is constant, so HEAT  = fk*D, where D is the distance moved and  fk = 17,000 (N) is  given. Note the initial kinetic energy KEi  is not zero as in problem 9.  Set the lowest point equal to the zero of gravitational potential energy (ie, y = 0 at point 2.)  Note U = Ug + Us, where Ug =  mgy and Us  = (1/2)*k*y'2,  where y' is the spring coordinate indicating compression. We have KEi  = 16,000 J and Ui = mgyi  + 0  since the initial spring potential  energy is zero (not compressed--- yi' = 0) ; note  yi = 2.00 m.   Uf = mgyf +  (1/2)*k*yf'2 , where yf = 1.00 m and 
yf' = -1.00 m. (Since you square y' in general the  negative sign goes away. ) The heat of course is   fk*D, where D = 1.00 m. Solve for  KEf  and find  final speed vf : 

KEi + Ui  = KEf + Uf + HEAT 
means 16,000 J + mgyi = KEf +  mgyf +  (1/2)*k*yf'2 + 17,000 (N)*d , where yf  =  d = 1.00 m and 
yf' = -1.00 m.

 
(b) If you think the acceleration is constant think twice; the spring makes things more complex as we  have already seen in Ch. 14.   So use:  sum of the forces in the y direction = m*a = pos - neg, where we set the positive y direction to be up. Thus: : m*a = | kyf' | +  fk - mg,  where clearly  the spring  and friction forces point upward as the object moves down. 
38.*   READ section 7.4 and 7.5:  If U  = U(x) then Fx = -dU/dx. The x-component of force is the negative  derivative of U; Thus if dU/dx is positive , the force points in the negative  x -direction (Fx < 0) and if dU/dx is negative then  force points in  positive x direction (Fx > 0.) When dU/dx = 0   force is zero and the object is said to be in equilibrium,  either stable or unstable. A CONCAVE UP POTENTIAL ENERGY FUNCTION MEANS EQUILIBRIUM IS STABLE.
41*.  (a) CHAPTER 5 REVIEW: You must prove whether or not the system is at rest; the emphasis here is the use of  Ch. 5 concepts, whereas Ch. 7 is more directly related  to part (b). 

If the system is at rest  and remains so we have for the concrete m*a = pos - neg = mg - T = 0 . So you can immediately find T. 

For the Box we have m'*a = 0   = T - fs ' , where fs ' is the force of static friction from the ground and m' is the Box mass.  We also  have m''*a = 0,  where m'' is the gravel mass. ( i.e. There is no friction force between the gravel and Box if the system is at rest.)  Compare  T and fsmax = us*N, where N is the magnitude of the normal force on the box, which must be equal to (m' + m'')*g.  

If T < fsmax ', then the system is at rest and remains so  and you can find the friction force  of magnitude fs ' on Box. 

(b)  Before you begin this part, you should check if the  system will move under this new condition. Assume the system is at rest and will remain so;  if you get a contradiction  then the assumption  is wrong--- this is called proof by contradiction, common in math classes. If the system is at rest and remains so, then fs ' = T < fsmax  ' .  When the gravel is removed, the maximum friction force magnitude now is fsmax ' = us*m'*g , now smaller.    

But T = m*g  = cement weight magnitude >  fsmax ', contradicting the assumption. The system won't remain  at rest and you can now use :

 KEi + Ui  = KEf + Uf + HEAT, where  KEi    = 0 and Ui = m*g*yi, where yi = 2.00 m.  (The lowest point for the cement in this problem is considered to be the zero of gravitational potential energy; note we did not include the gravitational potential  of the Box since it does not change. If we included it, it would cancel out from both sides of the equation.) 
HEAT = fk*D , where fk is the friction force magnitude on  Box given by CH. 5 THEORY: f = u*N, where u is the coefficient of kinetic friction.   Clearly the vertical drop distance of cement  equals the horizontal distance moved by Box. 
42.* KEi + Ui  = KEf + Uf + HEAT , where HEAT = 0. 

(a) See VIRTUAL LAB 2, question 1. 
(b)  The simplest way to do the problem is to bypass the speed and kinetic energy just after the Block  leaves the spring and consider only two points in space-time: When the spring is fully compressed and the block is at rest (i) and when the  block has  reached its maximum height at the top and is momentarily at rest before turning around and sliding back down (f). 
KEi + Ui  = KEf + Uf    . You should be able to see which terms are clearly zero. Note: Let the horizontal  level be defined  y = 0 and  the zero of gravitational potential energy.  Thus we have initial (i) spring potential energy = final (f) gravitational potential energy at the top .  Once you get the final potential energy at top  find the vertical height  H and use trig  to find the distance along the incline. 
43.* The simplest way to do the problem is to bypass the speed and kinetic energy just after the Block  leaves the spring and consider only two points in space-time: When the spring is fully compressed and the block is at rest (i) and when the  block has  reached its maximum horizontal distance  and comes to rest permanently (f) .
 KEi + Ui  = KEf + Uf + HEAT , where Ug = 0 at both points (ground level, y = 0) . You should be able to see which terms are clearly zero. Thus we have initial (i) spring potential energy = HEAT , where HEAT =   fk*D  :

(1/2)*k*d2 = u*mg*D,  where u =  coefficient of kinetic friction and d and D are given in diagram.  
46.* ANOTHER CLASSIC,  like #24. This one mixes energy conservation and dynamics just like #9 above and #63 below. 
(a) KEi + Ui  = KEf + Uf + HEAT , where HEAT = 0.
ENERGY CONSERVATION: (I) KEi + Ui  = KEf + Uf   , WHERE KEi = 0,  Ui = m*g*h, KEf  = (1/2)*m*v2  and Uf = m*g*2R: 

mgh =   (1/2)*m*v2   +  m*g*2R    AT TOP. 

CIRCULAR  DYNAMICS: ALSO AT TOP,  the normal force and the weight of magnitude mg point down in the pos direction toward the center of the circle.   Write:
(II) Sum of forces in radial  direction = m*v2/R = pos - neg = N + mg - 0 =  N + mg, since there is no force pointing in the neg direction away from the center; see #120 AND #42, Ch. 5 , point B at top.  

Set N = 0 IN THIS CASE.

Solve equations I and II simultaneously  for h in terms of R by eliminating v.  The mass m cancels out.

(b) FOR THIS INITIAL HEIGHT h,  greater than that of part (a), the object never loses contact with the track at the top so it does  reach point C. 

(I) ENERGY CONSERVATION :
(I) KEi + Ui  = KEf + Uf   , WHERE KEi = 0,  Ui = m*g*h, h = 3.50*R,   KEf  = (1/2)*m*v2  and Uf = 
m*g*R (at point C) .  Find v  at C:

mg*( 3.50*R )  =   (1/2)*m*v2   +  m*g*R

(II) CIRCULAR  DYNAMICS: At C,  the normal force points right in the pos direction toward the center of the circle.  The weight of magnitude points vertically down and does not contribute to the centripetal force. However, the weight does contribute to the tangential  acceleration since the weight at C is tangent to the circle.   Write:
(II) For the centripetal (center seeking) force, sum of forces in radial direction  = m*v2/R = pos - neg = N  - 0 =  N, since there is no force pointing in the neg direction away from the center. Note, at C  the radial acceleration is v2/R .  You cannot find N without the mass, not given.

(III) For the tangential acceleration, let the pos direction be vertically down, a direction tangent to the circle at point C. 
Sum of forces in tangential  direction = m*at =  mg - 0 = mg, where at is the tangential acceleration; m cancels  out. 
DRAW RADIAL AND TANGENTIAL acceleration components to scale.

47.* ANOTHER CLASSIC: m*g*H = TOTAL HEAT, SINCE THE INITIAL AND FINAL kinetic energies are zero. The Heat is the total heat after possibly numerous trips across the flat section. TOTAL HEAT =
 fk*L, where L is the total distance moved on flat surface before coming to rest. To find the number of trips, divide  TOTAL HEAT by   fk*D, where D = 30 m exactly.  From this ratio you can find the  number of complete trips; from the fractional remainder, you can find exactly where the object comes to rest on the flat surface. If the ratio is less than one, then the object does not make a complete trip across the flat surface. 
50.*  KEi + Ui  = KEf + Uf + HEAT , where HEAT = 0. See  problem 68 hint below.

i = bottom.

f = top just before leaving horizontal surface 70 (m) above ground zero ( where Ug = 0)

Ui = 0.

Uf = mgyf, where yf = 70 (m)

KEi   =  (1/2)*m*vi2  .

KEf   =  (1/2)*m*vf2  .

GO BACK TO CHAPTER 3 , review horizontal launch problems and solve for the speed vf  at the top satisfying the given condition. Then find   vi  using conservation of energy. 

55.* KEi + Ui  = KEf + Uf + HEAT , where HEAT = 0.

 Let the zero of gravitational energy be the ground.

# Initially (i) the system is at REST and  Ui = the gravitational potential energy of the 12.0 kg clock at the given initial height. 

# Finally (f), the 12.0 kg block is about to hit the ground and the 4.0 kg block is at the same height that the 12.0 kg had initially.  Uf = the potential energy of the 4.0 kg block which is at height of 2.00 m above the ground. Both blocks are  moving at the same speed vf  even though they are moving in the opposite directions---use that  information to write KEf as the  sum of the kinetic energies of the two blocks.   

Solve for the final common speed  vf   of the two blocks.
56.* USE  KEi + Ui  + WF  = KEf + Uf + HEAT, where  U is the gravitational potential energy in this case, KE is the kinetic energy,   WF is the work done by the external thrust force and the HEAT is fk*D. The constant  friction force of magnitude fk = 500 (N) is given; D is the distance moved along  ramp. 

# Initially (i), the rocket is at REST and Ui = the gravitational energy at the initial vertical height H above the ground. Note  the height H has a simple relationship with D, the distance moved along the ramp; use simple geometry, the angle of 53 degrees,  and the properties of right triangles to write H in terms of  D.  

WF = F*D, where D is the distance moved along the ramp and F is the magnitude of the given external thrust force = 2000 (N) exactly. 

# Finally (f), Uf = 0, HEAT =    fk*D AND  KEf = (1/2)*m*vf2 , where the final speed is given.

Solve for D.
63.* ANOTHER CLASSIC: THE STRUCTURE OF THIS PROBLEM,  LIKE  9, and 46,  mixes ENERGY CONSERVATION (Ch. 7) and CIRCULAR  DYNAMICS  (Ch. 5)

ENERGY CONSERVATION (Ch. 7):

 (I) KEi + Ui  = KEf + Uf + HEAT , where HEAT = 0.

# Initially (i), skier is at the top and is essentially at rest;  KEi      = 0.  Ui = m*g*R if the zero of potential  energy is considered to be at the level of the snow ball's  horizontal diameter. 
(I) mg*R = (1/2)*m*vf2  + mg*R*cos alpha, where alpha is angle with vertical shown. 
# Finally (f), we consider the point when the skier leaves the snow ball , i.e. when the normal of magnitude N becomes zero.   See figure 7.63. We define the location of this point in the following way: Draw a radial line from the snow ball center to this point and  define the angle with the vertical as alpha .  KEf  = (1/2)*m*vf2      and Uf = m*g*R*cos alpha.

CIRCULAR  DYNAMICS: At any point on the surface of the ball,  the normal force points radially away from the   center of the circle.  The gravitational force  of magnitude mg points vertically down and the component of the gravitational  force along the radial line is mg*cos theta and points toward the center of the circle.    Write:
(II) For the centripetal (center seeking) force, Sum of forces in radial direction  = 
m*v2/R = pos - neg = mg*cos theta  - N.

For this problem set N = 0. Thus: 

 m*vf2/R =  mg*cos theta. 


Solve equations I and II simultaneously  for cos theta by eliminating v.  Compute theta by evaluating cos -1. Your angle will be  between 45 and 90 degrees.       
68.  WE DO THIS PROBLEM USING TWO STYLES--FROM CH. 6 AND 7.
See #50 above.  

CH. 6 STYLE:  
(i) (1/2)*m*vf2 - (1/2)*m*vi2   = total work = Wg + Wf + WN,  work by gravity, the friction force and normal force respectively where  last term is zero since  the normal work on the way up slope must be vanish because that force  is perpendicular  to the motion i.e.  makes an angle of 90 degrees. ALSO, THE SLOPE IS FRICTIONLESS MAKING  WORK OF  FRICTION Wf  = 0.

(ii) FIND THE SPEED AT THE TOP OF THE CLIFF USING (1/2)*m*vf2 - (1/2)*m*vi2   = total work = Wg  WHERE THE NET VERTICAL DISPLACEMENT IS FROM LOW TO HIGH.  IS GRAVITY WORK POSITIVE OR NEGATIVE? 

(ii) AFTER LEAVING CLIFF, DURING FREE FALL (i.e., PROJECTILE MOTION)  THE  gravitational  work will be positive for this high to low motion. IN THAT PHASE OF PROBLEM YOU CAN USE CH. 3 METHODS AS IN LAB 3.

HERE'S  HOW YOU DO IT CH. 7 STYLE:
FIND THE SPEED AT THE TOP OF THE CLIFF USING   (1/2)*m*vi2   =  (1/2)*m*vf2 + mgh.  THEN USE CH. 3 METHODS (PROJECTILE MOTION.) 
70 (a) KEi + Ui  = KEf + Uf + HEAT , where HEAT = 0. The block is released from rest so that should tell you the value of KEi . The maximum speed occurs when U is a minimum and that location is shown in figure 7.42. Uf = 0 assuming both springs are un-deformed and the block is in equilibrium.  Ui = sum of two spring potential energies: Spring 1 is stretched by 0.15 m and spring 2 is compressed by 0.15  m. To evaluate Ui, apply the formula (1/2)*k*x2 for the  potential  energy to both springs and add the two results. Solve for KEf and the maximum  speed  vf  of the  block WHICH OCCURS WHEN BOTH STRINGS ARE UN-DEFORMED. 
(b) The simplest way to do this part is to  assume the initial position (i) of the block is the same as in part (a) but the final location (f) is when Spring 1 is at maximum compression. When  Spring 1 is at maximum compression,  the block is at rest momentarily (i):  Ui  =  Uf   = sum of  two spring potential energies.  Spring 1 is compressed a distance |x'| and Spring 2 is stretched by the same |x'| .  Ui was evaluated in part (a).   Plug the symbol  |x'| into the formula for Uf and solve for |x'| numerically. 
75. * SELECTIVELY USE  KEi + Ui  + WF  = KEf + Uf  where  U is the potential energy due to the spring force in this case, KE is the kinetic energy and   WF is the work done by the external force of magnitude F. NO FRICTION SO HEAT = 0. 
The FULL equation  with WF only applies between initial point A and  point-B but we know the process continues beyond that.  The Block will continue to move beyond point-B and eventually come to rest. At that  point, the full distance from the origin  is the amplitude A of subsequent simple harmonic motion about the origin and would be described by the function Acoswt for t >0 where t=0 is when the block reaches  maximum distance from  origin---see Ch. 14 and virtual lab 2.. So that is your task---to find that distance . Here are some tips:
(a) Find the speed of the block at point-B using :
KEA + UA  + WF  = KEB + UB  where WF = F*D.  Note KEA = 0 and UA = 0. Also WF = F*D, D = 0.25 m and  UB = (1/2)*k*(0.25)2 in  Joules. Find the final kinetic energy  and final speed at B.  
(b) Find the additional distance before coming to rest using:
KEB + UB  = KEC + UC   . NOTE:  KEB  is known from the previous part and UB = (1/2)*k*(0.25)2
Also KEC  = 0 and UC = (1/2)*k*xC2  .  Find xC.  Then find  0.60 m -  xC .
Note:  KEA + UA  + WF  = KEC + UC  or, WF  = KEC + UC, where  WF = F*D, D = 0.25 m  and KEC = 0.