q6

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

11/13/2010 05:42:33 PM 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

m_Aa_x = pos - neg
m_Aa_x = F - f_k.
m_Aa_x = F - µN
m_Aa_x = F - µm_Ag

0 = F - µm_Ag

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) ma_x = 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 f_k = 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 W_g = +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 is 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.

NOTE IN THIS CASE WE ARE TALKING ABOUT THE WORK DONE ON THE BALL, WHICH IS NEGATIVE .
THE MAGNITUDE F IS EQUAL TO THE MAGNT5IDE OF THE FORCE ON THE BALL. FROM NEWTON'S THIRD LAW, F IS ALSO EQUAL TO THE MAGNITUDE OF THE FORCE OF THE BALL ON THE GLOVE.

22. Change in KE = net W = f*Dcos180.
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) mv_f² - (1/2)mv_i², where the final speed is zero; initial was computed above. NOTE THE WOERK DONE BY GRAVITY IS -mgh, negative num ber because we are going from low to high after the ball is released.

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 = W_T+ W_g .
W_T = work done by tension force.
Wg = work done gravitational force, i. e. by the weight of the load.
Now W_gis easy. Since you are moving from low to high, W_g= -mgh, where h is given. Note: The gravitational work is negative. Now for the work by the tension force:
W_T = (Tcos0)D = T·h.
(c) W_T was discussed in the previous part.
(d) W_g was discussed in part (b).
(e) Net work = Change in KE = (1/2)mv₂² - (1/2)mv₁²
W_T+ W_g= (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) PE_g = 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 = W_F = (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 W_g + W_F = 0. Now, W_g = -mgh, where h is given. What then is W_F ?

(c) The actual work W_F could be more than the previous answer if there is a positive change in KE. In that case we would have:
(1/2)mv_f² - (1/2)mv_i² = W_g + W_F > 0. In this case: (1/2)mv_f² - (1/2)mv_i² - W_g = W_F , or
(1/2)mv_f² - (1/2)mv_i² + mgh = W_F.

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 + PE_s1 = (1/2)mv₂² + mgy₂. + PE_s2 .
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. NOTE: PE_s2 = 0, since the spring is no longer in the picture in this sense: the spring is now longer deformed after the ball leaves the spring and rises in the air.
(a)(1/2)mv₁² + mgy₁ + PE_s1 = (1/2)mv₂² + mgy₂. + PE_s2 .
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. y_f = d. . Find the second speed at the release point, 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 already been solved for symbolically in terms of h, r and g above.   (There is no negatively directed force.)
Set N = 0, and solve for h in terms of r.

41. (1/2) mv_o² + PE_so = (1/2)mv² + PE_s , where the y-values are zero. Now,
the initial speed is zero, the initial potential energy PE_so = (1/2)kx_o² and the potential energy at a later time PE_s= (1/2)kx².

The total mechanical energy is PE_so . 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)mv_f² + mgy_f.
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 = -W_fk = the negative of the work done by friction.
Thus ,
W_fk = - 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 = W_F/time, where W_F = work by player. We assume the players moves up the stairs at constant speed v, so the change in kinetic energy is zero. Then the Net Work = 0. But Net Work = W_g + W_F, where W_g = work of gravity and W_F = work by player. Thus W_F= -W_g. Finally, since you are moving from low to high, W_g = -mgh. Make the correct substitutions to find the Power, where h = (140)sin32 meters and given the time.