REAL TEST 4
SAMPLE FINAL EXAM FOR AU 12
 QUIZ 7 (ANSWERS)
NOTES
SAMPLE TEST2
SAMPLE TEST3
READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE. 
CH. 7 
CH. 7 - Exercise/problems   (Work) 4, 6, 8, 10, Work and Kinetic Energy: 19, 22, 24, 78, Springs: 26, 27, Conservation of Energy without and with friction:  45, 80,  32, 33, 87, 49,  46, 48, 100, 91, 50, 54, 55, 56, 57, 59, 81, 82,  Power: 64, 67  
TURN IN: 6, 8( try 24)  , 19,  78, 26, 33, 45 (try 32) , 49 (try 87) ,  87, 54,  Re-do 78 using energy conservation, 82 (try 55, 56) , 64 (try 67)
* DISCUSSIONS PROVIDED. 
SELECTED DISCUSSIONS TO exercises/problems BELOW.
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.
DISCUSSIONS
6.(a) WF = F*cos 30 *D
(b) Wg =  - mgh if motion is from low to high. Wg = +mgh is motion is from high to low. What is  h in the case of horizontal motion?  
(c) WN = 0 Explain.
(d) Compute: WF  +  Wg+ WN  .
(e) Repeat if theta = 50 degrees
8. 
(a) Wfk  = -f*D = -0.40*mgcos53*D, where D = 2.00 m.
(b) Wg = +mgh = +mgD*sin 53.
(c) WN = 0
(d)Wfk  +  Wg+ WN  .
19. . (a) Change in KE = KEf  - KEi  = Wfk  = -f*D = -0.40*mg*D. FInd D. Note KEf = 0.  and KEi =

(1/2)*(mass)*Vi2 where the initial speed is given.
(b) Repeat if the initial speed was twice the speed given

(c) Energy is never lost or created , it is only transformed into different forms.

78.
(a)  Change in KE = Wg = +mgh since the motion is high to low. h is given.
Change in KE =  KEf  - KEi  .  Note KEi = 0.  and KEf =

(1/2)*(mass)*Vf2 speed.  Find the final speed.
(b) 

Change in KE = Wg + Wfk= +mgh + Wfk.  h is given.
Change in KE =  KEf  - KEi  .  Note KEi = 0.  and KEf =

(1/2)*(mass)*Vf2 .  The final speed is 62 m/s. Find Wfk . You answer will be negative.
Another way of doing the problems is conservation of energy. KEi + Ui = KEf + Uf + Heat.  KEi = 0 and

KEf =

(1/2)*(mass)*Vf2 speed.  The final speed is 62 m/s.  Ui = mgh and Uf = 0 since the person lands at ground zero (y = 0). Find Heat. Note: Your answer will be negative and will have the same magnitude as your answer to (a). See redo below.

26.
(a) Let F be the magnitude of the applied force. Thus F =k*D. where D = 0.192 m - 0.170 m. F = 25 (N)


(b) Ws = (1/2)kxi2 - (1/2)k*xf2  . Note: WF = Work by applied force =  -Ws = (1/2)kxf2 - (1/2)k*xi2  , where  = xi =  0 and xf = 0.022 m. Find work by applied force.
(c)  F =k*D where k is known from part (a) and F = 50(N). Find D and find the final length L = 0.170 m + D

33. (a) Conservation of energy KEi + Ui = KEf + Uf. , where KEi = 0 and Ui = (1/2)*k*xi2 = 11.5 J.
Uf = 0 since at that point the spring is undeformed (xf = 0.) Find KEf = 1/2)*(mass)*Vf2  and the final speed.
(b) The greatest acceleration occurs  when the spring is compressed initially. Fx = -kx. Find x by solving  Ui = (1/2)*k*xi2 = 11.5 J for the initial value of x.   Take the negative root. Note Fx = -kx. is positive. Then  use m*ax = -kx. Find ax   > 0.  
45. Use conservation of energy. MORE DISCUSSIONS TO COME. (1/2)*mvo2  + mgyi = (1/2)*m*vf2 + mgyf. Note the final value of y is zero and the initial value of y is h. The initial  speed is vo .  The answers do not seem to depend on the direction of the launch.
49. (1/2)*mvi2  + mgyi = (1/2)*m*vf2 + mgyf. Here, the initial speed is zero. Find the final speed.  Note:
 y =  L*(1- cos theta), where L = 20 m. Get the initial value of y by setting theta = 45 degrees  and the final value of y by setting theta = 30 degrees.
87. This is a great problem incorporating many ideas, some new , some old (drawing upon ideas of circular dynamics in Ch. 6)
(a) (1/2)*mvi2 + mgyi = (1/2)*m*vf2 + mgyf ,  where the initial height is 25.0 m. and the firal height is 12.0 m; note the coaster starts from rest. 
(b) At B,  the positive direction is down in the direction of the centripetal force and  acceleration:
mvi2/R  += pos - neg = mg + N, where N is the  magnitude of the downward normal force on the coaster from the track. 
Solve for N. 
54. (a) Remember Us = (1/2)*kx2  =  3.20 J. Solve for x and take negative root to remind yourself it's compressed.
(b)  (1/2)*mvi2 + mgyi + (1/2)*kxi2= (1/2)*m*vf2 + mgyf   + (1/2)*kxi2   . 
The book starts from REST .  The initial height is 0.80 m from the top of spring and a distance 0.80 m + d from the position of the book after the book has compressed the spring to the maximum distance d,  when it momentarily comes to REST.
 mg(0.80 + d)  = (1/2)*kd2  . Solve this quadratic equation for d.
78.  Redo using KEi + Ui + KEf + Uf +HEAT, where Heat = f*h, where f is the magnitude of the force of air friction. 
(a) In this case f = 0 and  HEAT = 0, and the diver starts from rest; h is given.  Find the final kinetic energy and speed.
(b) In this case we start from rest, h is given and the final kinetic energy KEf is given through the final speed of 62 m/s.  Solve for f*h. Note you can get f if you want.
64. The motion is in the same direction  as the applied tension force of magnitude F. Power = F*v = 30.0 kW, where kW = 1000 W. Note v  given. Find F = tension force on motorboat from string.