TEST 3 FALL '11 SOLUTIONS BELOW; SCROLL DOWN WITHIN EACH PROBLEM TO SEE THEM. |
1.
(80 POINTS) A 85.0-kg
speed skier has finished a long down hill race and reaches a final slope (fig. 1 below) designed to slow her down. At the bottom of this slope her
speed is 29.0 m/s. She slides up the
inclined plane of snow on her skis and
at a certain vertical height h has speed 1.95 m/s. The force of friction
between her skis and the snow
does work of magnitude 3995.0 J . (Ignore
air friction.) (a) (44 point) What is the vertical height h? For full credit you must use the work – energy theorem or conservation of energy. (b) (8 points) What is the work Wg done by gravity during her trip up the inclined plane? Is this work positive or negative ? Give a brief explanation of the correct sign you choose (positive or negative.) (c) (4 points) What is the work done by the normal force ? Explain your answer. (d) (8 points) Use basic trigonometry to find the distance L moved along the incline. See fig. 2 for reference. (e) (16 points) Assume the force of friction between the skis and snow is constant with magnitude f. What is f? |
SOLUTIONS:
(a) (1/2)*mvi2 + mgyi
= (1/2)*m*vf2 + mgh + 3995.0 J, where the
initial y value is zero, the initial speed is 29.0 m/s and the
final speed 1.95 m/s. This gives h = 37.9 m. (b) Wg = -mgh = = -31600 J. This work is negative since the the gravitational force acts downward and the vertical component of displacement points up in the opposite direction. (c) ZERO since the normal force and motion are perpendicular (d) L = h/sin 30 = 75.8 (m) (d) -3995.0 J = -f*L, where L = 78.5 (m); thus f = 52.7(N). |
2. (80 POINTS) On a frictionless air track, a 0.5000 kg-glider moving to the right at 2.950 m/s collides with
a 0.4000 kg glider moving to the left at 1.950 m/s. The collision is elastic. Assume the right direction is the positive direction of motion . (b) (30 Points) What is the velocity of the
0.4000 kg-glider after the collision? (c) (8 points) Using the velocities of parts (a) and
(b), compute the total kinetic energy of the system after the collision. Do not round off during intermediate
computations.
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SOLUTIONS: (a) (0.500 kg)*(2.950 m/s) + (0.400 kg)*(-1.950 m/s) = (0.500 kg)*V + (0.400 kg)*V ' and 4.9 m/s = V' - V. V = -1.406 m/s and (b) V' = + 3.499 m/s. (c) (1/2)*(0.500 kg)*(-1.406 m/s)2 + (1/2)*(0.400 kg)*(-1.950 m/s)2 = 2.9360 J (d) (1/2)*(0.500 kg)*(2.950 m/s)2 + (1/2)*(0.400 kg)*(-1.950 m/s)2 = 2.9360 J (e) Answers are the same because collision is elastic . |
3.
EXTRA CREDIT. (20 POINTS) An airplane propeller is rotating at a steady 1900
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SOLUTIONS: (a) (1900)*(6.28)*(1/60) RAD/S = 199 RAD/S (b) THETA = 35*3.14/180 RAD = 0.610 RAD = 199*t; thus t = 0.003 seconds. (c) period = 1/f = 6.28/w = 6.28/198.97 seconds = 0.0316 seconds. (d) 18*(1/6.28)*( 60) RPM = 172 RPM. |
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