|REAL TEST 4|
|SAMPLE FINAL EXAM FOR AU 12|
|QUIZ 7 (ANSWERS)|
|READ ALL EXAMPLES. SOME PROBLEMS ARE JUST LIKE AN EXAMPLE.|
|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
ALL PROBLEMS ARE DUE EVEN THE ONES WITHOUT DISCUSSIONS, WHICH ARE DESIGNED TO HELP YOU DO RELATED EXERCISES.
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
(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.
(c) Energy is never lost or created , it is only transformed into different forms.
(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.
Change in KE = Wg + Wfk= +mgh + Wfk. h
(1/2)*(mass)*Vf2 . The final speed is 62 m/s.
Find Wfk . You answer will be negative.
(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.
(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*xi2 =
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
(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
(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.|