Quiz 6
Simulations Projectile Motion 
Ch.6: 1, 2, 4, 8, 19, 20, 22, 24, 25, 26, 27, 30, 31, 33, 34, 37, 38, 39, 40, 41, 43, 50, 58, 60, 67
turn ins : 4, 8, 22 (see #20), 24, 30, 34, 38, 41, 58, 60 
10/25/2009 10:56:28 AM Note correction to number 8(c).
HINTS POSTED BELOW; MEANWHILE SEE TEST SOLUTIONS AT THE BOTTOM OF THE HINTS BELOW. 
1. Wg > 0,  high to low 
2. Wg < 0,  low to high. 
4. The 

mAax = pos - neg
mAax = T - fk. 
mAax = T - µN
mAax = F - µmAg

0 = F - µmAg

Find F.
The find the work using equation  6.1.  It should be obvious  what is the angle is between the applied force of magnitude F and the motion .

Note by equation  6.1 the work done by the gravitational force and the normal  force are both zero since the horizontal motion is perpendicular   to both those forces.  
8. 
a) max = mgsin28 - µmgcos28 - F

where down the incline is positive.
 0 = mgsin28 - µmgcos28 - F, where F is the unknown magnitude of  force exerted by man.  
(b) Use  the magnitude F from part (a) and equation 6.1 , with  the angle at 180 degrees.  D = 3.6 m.  Work is negative.
(c) Use  the friction force magnitude  fk = µN = µmgcos28.  from Ch. 4  and equation 6.1, with angle 180 degrees,   D = 3.6 m.  Work is negative .
(d) The work done by gravity is Wg = +mgh since motion is from high to low. To get h use trigonometry: h = Dsin28.

19. change in KE = net W = FDcos0.
Change in KE = (1/2) m v22 - (1/2)mv12,  where the first speed is zero; find the second. . Note: By  equation  6.1 the work done by the gravitational force and the normal  force are both zero since the horizontal motion is perpendicular   to both those forces.  
20.Change in KE = net W = FDcos180.
Change in KE = (1/2) m v22 - (1/2)mv12,  where the second speed is zero; first is given. Find F. 
22. Change in KE = net W = fDcos180.
Change in KE = (1/2) m v22 - (1/2)mv12,  where the second speed is zero; first is unknown. The friction  force coefficient  is  f = µmg.   
24 Change in KE = net W = FDcos0.
Change in KE = (1/2) m v22 - (1/2)mv12,  where the first  speed is zero; second is unknown which you must compute.  
After the object is released with the above  speed, it flies upward  an unknown distance h given by
-mgh = (1/2) mvf2 - (1/2)mvi2,  where the final   speed is zero; initial was computed above.  
25.
(a) 
ma = pos - neg
ma = T - mg.  Note: a = 0.160g = (0.160)( 9.8 m/s2)
Find the tension T for the given mass m.
(b) Net work = WT + Wg     .
WT = work done by tension force.
Wg = work done gravitational force, i. e.  by the weight of the load.
Now Wg  is easy. Since you are moving from low to high, Wg = -mgh, where h is given. Note: The gravitational work is negative. Now for the work by the tension force:
WT  = (Tcos0)D = T·h.
(c) WT was discussed in the previous part.
(d) Wg was discussed in  part  (b).
(e) Net work = Change in KE = (1/2)mv22 - (1/2)mv12
WT + Wg    = (1/2) mv22 - (1/2)mv12 , where the first speed = 0. Find the second speed.
26. PEs = (1/2)kx2  = 25  J, where spring constant k is given.
27. change in PE = mg(y2 - y1), where  y2 - y1 = 1.2 m.
30.
(a)  PEg = mgy.
If the origin ( y = 0) is at the ground, then y = 2.20 m.
(b) If the origin ( y = 0) is at the top of the head,  y = (2.20 - 1.60) m = 0.60 m.
(c) In each case we assume the book is lifted at constant velocity, so the net acceleration is zero.
Thus: ma = pos - neg
0 = F - mg , where F is the magnitude of the lift force exerted by the student. Thus F = mg.
The work by student = WF = (Fcos0)h = F·h = mgh, where h = full height through which the book is lifted; h = 2.20 m.
Thus , the  answer here matches the result of part (a) , not part (b).  
31.
(a) change in PE = mg(y2 - y1), where  y2 - y1 is given.
(b) Minimum work occurs at constant velocity, where the acceleration a = 0 on the way up.   
In this case, we have a similar situation as in the previous problem.
Another way of looking at this is by assuming the change in kinetic energy is zero; then the work by the person is a minimum.
 In that case,  change in KE = 0 = net work .

Thus Wg + WF = 0. Now,  Wg = -mgh, where h is given. What then is WF ?  

(c) The actual work WF   could be more than the previous answer if there is a positive change in KE. In that case we would have:
(1/2)mvf2 - (1/2)mvi2  =  Wg +  WF > 0.   In this case:  (1/2)mvf2 - (1/2)mvi2  -  Wg =  WF ,  or
(1/2)mvf2 - (1/2)mvi2  + mgh = WF .
33. Full disclosure: I  dislike Tarzan movies for their misrepresentations of African  cultures and peoples. But I 'd thought I'd assign this problem anyway, with that disclaimer, because of valuable  physics. Use conservation of energy: 
(1/2) mv12  + mgy1 =  (1/2)mv22 + mgy2.  

The first velocity of is given. Square it, use it. The first y value can be set to ZERO. The second velocity must be zero since Jane is at maximum height. Find the second y value. The length L of the vine does not affect your answer unless you wanted  the maximum angle made with the vertical. 
34. Skiing is a big pastime in CA, and low frictional forces are important to make the grade. Since the speeds in cross-country skiing are smaller than down hill,   friction is not as big a problem. But downhill needs low friction  to reach the speeds bringing  that legendary   "thrill" and "high, " and also a  gold medal   if you happen to be  competing in the Winter Olympics. 

(1/2) mv12  + mgy1 =  (1/2)mv22 + mgy2.  The first velocity of is ZERO.  The first y value is given.   The second velocity is sought. The second y value is ZERO. Note: the angle is unnecessary because there is no friction. 
37. 

(a) (1/2) mv12  + mgy1 =  (1/2)mv22 + mgy2  , where the first velocity is given and the first y-value is 3.0 m. The second y - value is ZERO, at the level of the un-depressed trampoline.  Find the second speed by taking the plus root of the solution. The negative root gives you the downward velocity if UP is positive.    
(b)  Now you can set the y level of the depressed trampoline to be zero. The kinetic and potential energy of the artist just before impact is transformed into the potential  energy of the trampoline spring:
(1/2)mv2 + mgd  = (1/2)kd2,  where d is the sought for  distance the spring is compressed when the artist comes to rest momentarily.  Note : v is the speed you found in part (a) . You must now solve a quadratic equation to find d. 
38. (1/2) mv22  + mgy1 =  (1/2)mv22 + mgy2. The first speed is given; note that the angle is not needed. The first y-value is given as 265 m, and the second y-value is ZERO.  
39. I thought I'd do part (b) first. 
(b) (1/2)mv12  + mgy1 + PEs1 =  (1/2)mv22 + mgy2. + PEs2 . 
The first values of the speed and y are ZERO. The second speed value is ZERO. 
The first spring potential energy value is (1/2)kd2 , where d is the given distance the spring is initially compressed.  Find the second  y-value y2  relative to ZERO. 
(a) (1/2) mvi2  + mgyi + PEsi =  (1/2)mvf2 + mgyf. + PEsf . 
The first (initial) values of the speed and y are ZERO. You seek the  final speed value when the mass is at the level of its release. The first spring potential energy value is (1/2)kd2 , where d is the given distance the spring is initially compressed.  The second y- value is the given d; i.e. yf = d.  The second spring potential energy is ZERO. Find the second (final)  speed, evaluated at the level where the spring is un-deformed, the second spring potential energy is ZERO, and  the mass leaves the platform at the spring's end.
40. 
 (1/2)mv12  + mgy1 =  (1/2)mv22 + mgy2. 
The first speed is ZERO and the first y- value is h. The second y value is 2r. Solve for the second speed symbolically in terms of h,  r and g.   Note the mass cancels out. 

Now, at the top , the net force is down, toward the center of the circle. The net force is
 mv22/r = pos - neg  (Review Chapter 5.)   
mv22/r = N + mg - 0 , where the second speed has been solved for symbolically in terms of h, r and g.   (There is no negatively directed  force.)  
Set N = 0, and solve for h in terms of r.
41. (1/2) mvo2  +  PEso =  (1/2)mv2 +  PEs , where the y-values are zero. Now, 
the initial speed is zero,  the initial potential energy PEso  = (1/2)kxo2  and the potential energy at a later time PEs = (1/2)kx2.  

The total mechanical energy is  PEso  .  The total mechanical energy can also be written as the sum on the right hand side of the above equation; either expression represents the total mechanical energy. 
43. Let the y-value at point 2, the lowest point,  be zero .   Thus y1 = 35 m. Note: initial ( first) speed is zero. 
(1/2) mv12  + mgy1 =  (1/2)mvf2 + mgyf. 
We have 3 cases where f = 2, 3 and or 4, where the y value is  given in each case.  Find each speed. 
50. 

(a) (1/2) mv12  + mgy1 =  (1/2)mv22 + mgy2.  The first velocity of is ZERO.  The first y value is given as as h = 13.0 m.    The second velocity is sought. The second y value is ZERO.  

(b) (1/2) mv12  + mgy1 =  (1/2)mv22 + mgy2  + HEAT   The first velocity of is ZERO.  The first y value is given as h =
 13.0 m.    The second velocity is given. The second y value is ZERO. Solve for HEAT. Then, remember that HEAT = -Wfk = the negative of the work done by friction. 
Thus ,  
Wfk = - HEAT.  
-f·h = -HEAT, where f = magnitude of friction force.  Thus:

f·h = HEAT.

Find f.  
58. Power = Work/time = FD/time, where F = magnitude of force exerted by motor on piano.   We assume the piano moves upward at constant speed v, so the  upward acceleration is zero, thus 0 = F - mg, so F = mg.  D is given. The Power is given.  Solve for time. 
60. Average Power = Work/time = F ·D/ time , where F =  force. Solve for the force using F = ma, where a = acceleration. You can get acceleration a using  Chapter 2 methods and the given time and final speed.( Convert to m/s.)  Also D is obtained from the Ch. 2 equation:
D = (1/2)a·t2 .  

Note: An easier way to solve the problem is to write Work as the change in kinetic energy:

Work = (1/2)mv22 - (1/2)mv12  , where first speed is zero, and second speed is given. Thus,
Average Power = Work/time =  [(1/2)mv22 - (1/2)mv12]/time. 
67. Power = WF/time, where  WF = work by player. We assume the piano moves upward at constant speed v, so the change in kinetic energy is zero. Then the Net Work = 0. But Net Work = Wg + WF, where Wg = work of gravity and WF = work by player. Thus  WF = -Wg.  Finally, since you are moving from low to high,  Wg = -mgh. Make the correct substitutions to find the Power, where h = (140)sin32 meters and given the time.  
more hints later.
Test 1 Solutions Sec 01 followed by Sec 071 (Night Class) 
Sec.  01 Scantron:  aabaaaabba
1. 90 km/h = 90)(5/18) = 25.0 m/s. 
(a) -4.0 m/s2 = (0 - 25.0 m/s)/(change in t).   Thus (change in t) = 6.3 seconds. 
(b) average V = (Vf + 25.0 m/s)/2 = (0 + 25.0 m/s)/2 = 12.5 m/s 
(c) 
Displacement for the first 1.5 seconds: (25 m/s)( 1.5 s ) = 37.15 (m)
Displacement for the last 6.3 seconds: Use Vf2 = 0 = (25 m/s)2 + 2(-4 m/s2)(change in x) . Thus (change in x) = 78.125 (m) 

Note that so far the number of significant figures is greater than the required, but these are intermediate computations. We will round down when we add the the two displacements: 37.15 (m) + 78.125 (m) = 115.275 = 120 (m)
(d) 115.275 (m) /(1.5 + 6.25)s = 15 m/s
2. 
(a) Let down be positive.  In the net,  acceleration a = (change in velocity)/(change in time). We need two things,  a and the change in velocity in order to get the (change in time) in net. 

The final velocity at the very bottom is zero, and the velocity just before the jumper hits the net can be easily found using: V22 = 2g(16 m) = 313. 6 m2/s2. Thus V2 = 17.7 m/s. Now find the acceleration a: 02 = V22 + 2a(1.2 m), leading to 
a = -130 m/s2.   Now we can get the change in time:
-130 m/s2 = (0 - 17.7 m/s)/(change in time) leads to (change in time) = 0.135 seconds. 
(b) The average velocity = total displacement / total time = (16 + 1.2)/(total time). Now, the total time must include the time interval  in the air before hitting the net: 16 = (1/2)gt2 leads to t  = 1.8 seconds

THUS AVERAGE VELOCITY = (17.2 m)/(1.94 s) = 8.86 m/s, where   the denominator is the sum of both times.  
3. V12sin2Ø/g = 7 where V1 = 100 m2/s2 . Thus: sin2Ø  = 0.686, which has two solutions. 

2Ø = 43.31 degrees in quadrant 1 and  2Ø= 180 -  43.31 = 136.69 in quadrant 2. Quadrant 1 and 2 are where the sine function  has the same positive value. 
Thus Ø = 21.66 degrees and 68.34 degrees after dividing by 2 in both cases. 
(a) 21.66 degrees 
(b) When  the height is a maximum, the y-component of velocity is zero ( at the top) . So write:
0 = (V1sin21.66)2 - 2gh, where h = max height = 0.695 (m) 
(c) 68.34 degrees 
(d) 0 = (V1sin68.34)2 - 2gh, where h = max height =  4.4(m) 
(e) Let  V1 = 9.5 m/s after being reduced by 5 %. Now find :
V12sin2Ø/g = 6.3 (m) so the change in height is 0.7 (m) . 
4. 
(a)
Ax = 3.6cos60 = 1.8
Ay = 3.6sin60  = 3.1
Bx = -2.4cos30 = -2.1
By = -2.4sin30  = -1.2
(b) C2 = Cx2 + Cy2 

where C = magnitude; the components are being squared and summed.  

Now 

Cx  = - 0.3 

Cy =  1.9 , after adding the corresponding components .

C = 1.92 after evaluating the positive square root. 
(c) Since the x-component of the resultant is negative and the y- component is positive, the resultant lies in the 2nd quadrant. 

Note: tanØR = 1.9/0.3, where ØR  is the related angle  in the second quadrant,  81 degrees. 
Test 1 Solutions  Sec 071 (Night Class) Scantron: abbbccabbbb
1. Convert 90 km/h and 70 km/h to 25 m/s  and 19.44 m/s,  respectively by multiplying by conversion  factor  5/18.  
 (a) 1200 - 19.44t = 25t gives t = 27 seconds. Thus car distance = (25 m/s)t = 675 (m). 
(b)  truck distance = (19.44 m/s)t = 525 (m). 
(c) 27 seconds from part (a). 
2. 
(a) Let down be positive.  In the net,  acceleration a = (change in velocity)/(change in time). We need two things,  a and the change in velocity in order to get the (change in time) in net. 

The final velocity at the very bottom is zero, and the velocity just before the jumper hits the net can be easily found using: V22 =(5 m/s )2 +  2g(13 m) =279.0  m2/s2. Thus V2 =16.72 m/s. Now find the acceleration a: 02 = V22 + 2a(1.1 m), leading to a = -127 m/s2.   Now we can get the change in time:
-127 m/s2 = (0 - 16.72 m/s)/(change in time) leads to (change in time) = 0.131 seconds. 


(b) average V = (Vf + 25.0 m/s)/2 = (0 + 16.727 m/s)/2 = 8.4 m/s 
Or,  average V =(1.1 m)/(0.131 s)  = 8.4 m/s 

(c) The average velocity = total displacement / total time = (13 + 1.1)/(total time). Now, the total time must include the time interval  in the air before hitting the net: 13 =5t +  (1/2)gt2 leads to t  = 1.19 seconds. You could also get the time t in this way:

g = (change in velocity) / (change in t) = (16.72 - 5)( change in t) leads to (change in t) between the top and net =  1.19 seconds (same answer) . 

THUS AVERAGE VELOCITY = (14.1 m)/(1.33 s) = 10.62 m/s, where  the denominator is the sum of both  times.  
3. V12sin90/g = 7.  Solve for  V1 = 8.28  m/s . 

(a) When  the height is a maximum, the y-component of velocity is zero ( at the top) . So write:
0 = (V1sin45)2 - 2gh, where h = max height = 1.75 (m) 
(b) Let V1 = 7.78 m/s after being reduced by 6 %. Now find :
V12sin90/g = 6.18 (m) so the change in height is 0.82 (m) . 
4. 
(a)
Ax = 3.6cos60 = 1.8
Ay = 3.6sin60  = 3.1
Bx = -2.4cos30 = -2.1
By = -2.4sin30  = -1.2
(b) C2 = Cx2 + Cy2 

where C = magnitude; the components are being squared and summed. 

Now 

Cx  = - 0.3 

Cy =  1.9 , after adding the corresponding components .

C = 1.92 after evaluating the positive square root. 
(c) Since the x-component of the resultant is negative and the y- component is positive, the resultant lies in the 2nd quadrant. 

Note: tanØR = 1.9/0.3, where ØR  is the related angle  in the second quadrant,  81 degrees.