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, Redo 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) W_{F} = F*cos 30 *D (b) W_{g} =  mgh if motion is from low to high. W_{g} = +mgh is motion is from high to low. What is h in the case of horizontal motion? (c) W_{N} = 0 Explain. (d) Compute: W_{F} + W_{g}+ W_{N} . (e) Repeat if theta = 50 degrees 
8. (a) W_{fk} = f*D = 0.40*mgcos53*D, where D = 2.00 m. (b) Wg = +mgh = +mgD*sin 53. (c) W_{N} = 0 (d)W_{fk} + W_{g}+ W_{N} . 
19. . (a) Change in KE = KEf 
KEi = W_{fk} = f*D = 0.40*mg*D. FInd D. Note KEf =
0. and KEi =
(1/2)*(mass)*Vi^{2} where the initial speed is 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)*Vf^{2} speed. Find the final speed. Change in KE = Wg + W_{fk}= +mgh + W_{fk}. h
is given. (1/2)*(mass)*Vf^{2} . The final speed is 62 m/s.
Find W_{fk} . You answer will be negative. KEf = (1/2)*(mass)*Vf^{2} 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)

33. (a) Conservation of energy KEi + Ui =
KEf + Uf. , where KEi = 0 and Ui = (1/2)*k*x_{i}^{2} =
11.5 J. Uf = 0 since at that point the spring is undeformed (x_{f} = 0.) Find KEf = 1/2)*(mass)*Vf^{2} and the final speed. (b) The greatest acceleration occurs when the spring is compressed initially. F_{x} = kx. Find x by solving Ui = (1/2)*k*x_{i}^{2} = 11.5 J for the initial value of x. Take the negative root. Note F_{x} = kx. is positive. Then use m*a_{x} = kx. Find a_{x} > 0. 
45. Use conservation of energy. MORE DISCUSSIONS TO COME. (1/2)*mv_{o}2 + mgy_{i} = (1/2)*m*v_{f}^{2} + mgy_{f}. Note the final value of y is zero and the initial value of y is h. The initial speed is v_{o} . The answers do not seem to depend on the direction of the launch. 
49. (1/2)*mv_{i}^{2} + mgy_{i}
= (1/2)*m*v_{f}^{2} + mgy_{f}. 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)*mv_{i}^{2} + mgy_{i} = (1/2)*m*v_{f}^{2} + mgy_{f }, 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: mv_{i}2/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)*kx^{2}
= 3.20 J. Solve for x and take negative root to remind yourself
it's compressed. (b) (1/2)*mv_{i}^{2} + mgy_{i} + (1/2)*kx_{i}^{2}= (1/2)*m*v_{f}^{2} + mgy_{f } + (1/2)*kx_{i}^{2 } . 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)*kd^{2 } . 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. 