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course phy 201
11/10 3:20 am
`q001. A velocity of 5 m/s at 30 degrees is added to a velocity of 1 m/s at 120 degrees, What is the resulting velocity?****
vX: 5 m/s, 30 degrees; vY: 1 m/s, 20 degrees; horizontal velocity vX: 5 * cos(30 degrees) = 4.33 m/s; vertical velocity vX: 5 * sin(30 degrees) = 2.5 m/s; horizontal velocity vY: 1 * cos 120 degrees = -0.5 m/s; vertical velocity vY: 1 * sin(120 degrees) = 0.866 m/s; total horizontal velocity: 3.83 m/s; total vertical velocity: 3.366 m/s; magnitude of final vector: roughly 5.1 m/s; angle of final vector: arctan(3.366/3.83) = roghly 41.33 degrees;
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`q002. Sketch, reasonably to scale, the force vectors acting on a system of two masses, one of mass 10 kg on a horizontal frictionless surface and the other of 2 kg suspended by a string over a pulley, with the other end of the string connected to the 10 kg mass. Between the 10 kg and the mass the string is horizontal. The tension in the string is considered internal to the system; the tension pulls one way on the 10 kg mass and the other way on the 2 kg mass, but you don't need to sketch the tension vector.
How many force vectors did you sketch?
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four
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Identify each of these force vectors and explain what the source of each force is.
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The major force vector (v1) is pointing strait down from the 2 kg mass. This is caused by gravity. The second vector (v2) is pointing directly down by the 10 kg mass. This is gravity’s force on the larger mass. The third vector (v3) is pointing from the 10 kg mass, away from the pulley. This is friction. The final vector (v4) is pointing strait up from the 10 kg mass. Since the surface under the 10 kg mass isn’t breaking, then it must be exerting a force of equal mass to stop it.
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List the vectors in order, from the smallest to the largest.
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v3, v2, v4, v3
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What is the magnitude of each of your vectors, as a percent of the largest?
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v2: 20%
v4: 100%
v3: 0.4% (assuming 2% friction)
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What is the magnitude of the net force on the system as a percent of the largest?
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v3/v4 cancel each other out. Friction works against 2kg acting. Total percentage: 19.6%
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Sketch to the same scale the force gravity would exert on the two masses, were combined into one. What is the magnitude of the net force on the system as a percent of this force?
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120%
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`q003. Sketch the force vectors acting on a system of two masses, one of mass 10 kg and the other of 8 kg with the masses suspended on opposite sides by a string over a light frictionless pulley. The tension in the string is considered internal to the system, and the tension pulls each mass toward the other.
How many force vectors did you sketch?
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four
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Identify each of these force vectors and explain what the source of each force is.
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There are two vectors pointing strait up, and two pointing strait down, one from each of the masses. These are the force of gravity on each of the masses, and the resulting pull on the other mass because of it.
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List the vectors in order, from the smallest to the largest.
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10 kg up (v1)
8 kg down(v2)
10 kg down (v3)
8 kg up (v4)
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What is the magnitude of each of your vectors, as a percent of the largest?
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v1 = v2
v3 = v4
v1/v2 are 80% as great as v3/v4
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What is the magnitude of the net force acting on the system, as a percent of the largest?
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20%
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Sketch to the same scale the force gravity would exert on the two masses, were combined into one. What is the magnitude of the net force on the system as a percent of this force?
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180%
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Will the acceleration of the system be greater or less than the acceleration of gravity?
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less
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`q004. If you were 10 000 km from the center of the Earth, the gravitational attraction between you and the Earth would be (6350 km / (10 000 km) ) ^ 2 times as great as it is on the surface of the Earth (6350 km is the radius of the Earth, so (6350 km / (10 000 km) ) is the inverse ratio of the 10 000 km distance to your distance from the center).
Assume that you and your spacesuit have a mass of 100 kg. What gravitational force would you experience at the 10 000 km distance?
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ratio = 0.403; acceleration on surface: 9.8 m/s^2; acceleration at 10,000 km; 3.95 m/s^2; force on 100 kg suit: 395 Newtons
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What would be your acceleration toward the center of the Earth, assuming no other forces to be present?
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yes
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If you were in a circular orbit your centripetal acceleration as you move around the circle was equal to the gravitational acceleration you just calculated. The centripetal acceleration of a point moving at speed v around a circle of radius r is v^2 / r. What would be your velocity, were you in a circular orbit?
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3.95 m/s^2 = v^2/10,000km; 10,000 km = 10,000,000 meters; 3.95 m/s^2 = v^2/10,000,000 meters; 39,500,000 m^2/s^2 = v^2; 6284.9 m/s = v; velocity = roughly 6284 m/s;
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course phy 201
11/10 3:20 am
`q001. A velocity of 5 m/s at 30 degrees is added to a velocity of 1 m/s at 120 degrees, What is the resulting velocity?****
vX: 5 m/s, 30 degrees; vY: 1 m/s, 20 degrees; horizontal velocity vX: 5 * cos(30 degrees) = 4.33 m/s; vertical velocity vX: 5 * sin(30 degrees) = 2.5 m/s; horizontal velocity vY: 1 * cos 120 degrees = -0.5 m/s; vertical velocity vY: 1 * sin(120 degrees) = 0.866 m/s; total horizontal velocity: 3.83 m/s; total vertical velocity: 3.366 m/s; magnitude of final vector: roughly 5.1 m/s; angle of final vector: arctan(3.366/3.83) = roghly 41.33 degrees;
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`q002. Sketch, reasonably to scale, the force vectors acting on a system of two masses, one of mass 10 kg on a horizontal frictionless surface and the other of 2 kg suspended by a string over a pulley, with the other end of the string connected to the 10 kg mass. Between the 10 kg and the mass the string is horizontal. The tension in the string is considered internal to the system; the tension pulls one way on the 10 kg mass and the other way on the 2 kg mass, but you don't need to sketch the tension vector.
How many force vectors did you sketch?
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four
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Identify each of these force vectors and explain what the source of each force is.
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The major force vector (v1) is pointing strait down from the 2 kg mass. This is caused by gravity. The second vector (v2) is pointing directly down by the 10 kg mass. This is gravity’s force on the larger mass. The third vector (v3) is pointing from the 10 kg mass, away from the pulley. This is friction.
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Good. However in this case the surface is frictionless, so this force would be zero.
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The final vector (v4) is pointing strait up from the 10 kg mass. Since the surface under the 10 kg mass isn’t breaking, then it must be exerting a force of equal mass to stop it.
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List the vectors in order, from the smallest to the largest.
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v3, v2, v4, v3
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What is the magnitude of each of your vectors, as a percent of the largest?
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v2: 20%
v4: 100%
v3: 0.4% (assuming 2% friction)
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What is the magnitude of the net force on the system as a percent of the largest?
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v3/v4 cancel each other out. Friction works against 2kg acting. Total percentage: 19.6%
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Sketch to the same scale the force gravity would exert on the two masses, were combined into one. What is the magnitude of the net force on the system as a percent of this force?
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120%
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`q003. Sketch the force vectors acting on a system of two masses, one of mass 10 kg and the other of 8 kg with the masses suspended on opposite sides by a string over a light frictionless pulley. The tension in the string is considered internal to the system, and the tension pulls each mass toward the other.
How many force vectors did you sketch?
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four
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Identify each of these force vectors and explain what the source of each force is.
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There are two vectors pointing strait up, and two pointing strait down, one from each of the masses. These are the force of gravity on each of the masses, and the resulting pull on the other mass because of it.
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You're better off thinking in terms of string tension than the force exerted by one mass on the other. The string tension in this case will be the same on both masses (if the pulley had mass or if a frictional force was tending to work against its rotation, the tension in one side of the string would differ from that in the other).
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List the vectors in order, from the smallest to the largest.
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10 kg up (v1)
8 kg down(v2)
10 kg down (v3)
8 kg up (v4)
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The upward forces would be equal, both due to the constant tension in the string.
If the forces were as you list them the system would be in equilibrium.
It's good to think of the tension, but for present purposes the tension can be regarded as internal to the system.
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Note also that forces aren't measured in kg.
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What is the magnitude of each of your vectors, as a percent of the largest?
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v1 = v2
v3 = v4
v1/v2 are 80% as great as v3/v4
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What is the magnitude of the net force acting on the system, as a percent of the largest?
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20%
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Sketch to the same scale the force gravity would exert on the two masses, were combined into one. What is the magnitude of the net force on the system as a percent of this force?
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180%
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This is the present force as a percent of the largest force.
The net force is 20% of the largest force, the present force is 180% of the largest force.
What therefore is the net force as a percent of the present force?
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Will the acceleration of the system be greater or less than the acceleration of gravity?
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less
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`q004. If you were 10 000 km from the center of the Earth, the gravitational attraction between you and the Earth would be (6350 km / (10 000 km) ) ^ 2 times as great as it is on the surface of the Earth (6350 km is the radius of the Earth, so (6350 km / (10 000 km) ) is the inverse ratio of the 10 000 km distance to your distance from the center).
Assume that you and your spacesuit have a mass of 100 kg. What gravitational force would you experience at the 10 000 km distance?
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ratio = 0.403; acceleration on surface: 9.8 m/s^2; acceleration at 10,000 km; 3.95 m/s^2; force on 100 kg suit: 395 Newtons
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What would be your acceleration toward the center of the Earth, assuming no other forces to be present?
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yes
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What would be your acceleration?
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If you were in a circular orbit your centripetal acceleration as you move around the circle was equal to the gravitational acceleration you just calculated. The centripetal acceleration of a point moving at speed v around a circle of radius r is v^2 / r. What would be your velocity, were you in a circular orbit?
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3.95 m/s^2 = v^2/10,000km; 10,000 km = 10,000,000 meters; 3.95 m/s^2 = v^2/10,000,000 meters; 39,500,000 m^2/s^2 = v^2; 6284.9 m/s = v; velocity = roughly 6284 m/s;
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*#&!*#&!
&#Good responses. See my notes and let me know if you have questions. &#