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course phy 201
11/4 2:00 pmThere was a large gap between when I started work on this and when I finished it. Most notably, there may be some fairly obvious and stupid errors in problem 2 due to the break in my train of thought
`q001. An incline is angled at 5 degrees above horizontal. A 15 kg block slides along the incline.
If the xy axes are oriented so that the x axis is parallel to the incline, what are the x and y components of the block's weight?
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cos(265 degrees) * 15 kg = -1.3 kg;
sin(265 degrees) * 15 kg = -14.9 kg;
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The weight is a force, and therefore has a direction so you can calculate its components.
You can't calculate the components of a mass. Mass doesn't have direction.
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If the only forces acting perpendicular to the block are the y component of the weight and the normal force, then what is the normal force?
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normal force: 147 Newtons;
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If friction exerts a force whose magnitude is equal to 8% of the magnitude of the normal force, then what is the magnitude of the frictional force?
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11.76 newtons;
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You appear to have multiplied the 'mass components' (quotes because there is no such thing as mass components, since mass isn't a vector quantity) by the acceleration of gravity.
This does give you the correct force components, but you should have calculated the components based on the force in the first place.
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If the block is moving in the positive x direction, is the frictional force positive or negative (relative to the x direction)?
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positive;
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The frictional force opposes the relative motion of the object and the surface, so in this case it's negative.
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In this case what is the net force acting on the mass? (If you weren't able to figure out the frictional force, you can use 11 Newtons for the frictional force; that isn't quite right, so use the correct value if you got it.)
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horizontal force: -12.8
frictional force: 11.76
net force: -1.04 newtons
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If the object is moving in the positive direction (which would be up the incline; this isn't the direction in which it would travel of released from rest but it could easily have been given a push), then the frictional force is negative and the net force is about -25 Newtons.
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What therefore is its acceleration?
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-0.069 m/s^2;
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If the block is moving in the negative x direction, is the frictional force positive or negative (relative to the x direction)?
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negative
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The frictional force on the object will be in the direction opposite its motion.
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In this case what is the net force acting on the mass?
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1.04 newtons
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If both the parallel component of the weight and the frictional force were negative the net force would be about -25 N.
However that isn't the case here.
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What therefore is its acceleration?
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0.069 m/s^2
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What definitions, laws and procedures did you use in answering these questions?
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definition of force, force vectors
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You've done most of the steps correctly.
However this problem has an important point, which you missed because of the sign errors I pointed out.
You will need to rework this one, which should be easy and quick once you get the signs right.
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`q002. Suppose the block in the preceding slides 4 meters along the incline, in the positive x direction.
What work is done on it by the component of its weight parallel to the incline?
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Force: -12.8 newtons; ds: 4 meters; -51.2 Joules
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What work is done on it by the frictional force?
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11.76 newtons; 4 meters; 47 Joules
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Friction will do negative work.
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What is the change in its gravitational PE, and how much work is done on it by nonconservative forces?
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change in gravitational PE: 146.2 Joules; nonconservative forces: initial force, friction not included: -51.2 Joules
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The work done by gravity is -51.2 Joules, not 146.7 Joules.
The work done by a force is equal to the product of the component of the force in the direction of motion (in this case -13 N) and the displacement.
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What therefore is the change in its KE?
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51.2 Joules
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This would be the change in the KE of an object sliding down 4 m down the ramp, provided friction is absent.
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Answer the same questions if the block slides 4 meters in the negative x direction.
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-12.8 newtons * -4 meters: 51.2 joules; 11.76 newtons * -4 meters: - 11.7 joules; change in gravitational PE: 146.2 Joules; nonconservative forces: 51.2 joules
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See previous notes and see if you can answer this correctly. Many of your quantities are correctly calculated and are relevant, but not all.
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Is the magnitude of the KE change the same when the block slides in the positive direction as in the negative?
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yes
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Why do you think this is?
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the distance travelled and the force are exactly the same. This means that the work done is the same in magnitude, but in different directions. The change in KE is equal to the work done.
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Does the magnitude of the KE change depend on the initial velocity of the block?
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no
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What minimum KE would the block need to start with to travel at least 4 meters up the incline?
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51.2 joules
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What definitions, laws and procedures did you use in answering these questions?
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definition of KE, KE = work done, force vectors
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`q003. Masses of 30 kg and 31 kg are suspended from a light frictionless pulley by a cord of negligible mass.
If the system starts from rest, and upon release accelerates through a 5 meter displacement, by how much does the gravitational PE of the less massive block change?
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it increases by: 9.8 m/s^2 * 30 kg = force; * 5 meters = PE; 1470 joules;
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By how much does the gravitational PE of the more massive block change?
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decreases by: 9.8 * 31 kg = force; * 5 meters = PE; -1519 joules
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What therefore is the change in the PE of the entire system?
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-49 Joules
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What therefore is the change in the KE of the system?
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49 Joules
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What therefore must be its KE at the end of the 5 meter displacement?
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49 Joules
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What is the mass of the system?
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61 kg
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What therefore must be its velocity at the end of the 5 meter displacement?
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49 = ½ 61 * v^2; 49 = 30.5 * v^2; 1.6 = v^2; 1.26 = v; 1.26 m/s
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What is the net force acting on the system?
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force acting on heavy block - force acting on light block; 303.8 - 294 = 9.8 newtons;
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What therefore is the acceleration of the system?
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9.8 / 61 = 0.16 m/s^2;
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What velocity would the system attain with this acceleration, starting with initial velocity 0 and accelerating through the 5 meter displacement?
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vf^2 = v0^2 + 2a* ds; vf^2 = 0.32 * 5; vf^2 = 1.6; 1.26 m/s
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What definitions, laws and procedures did you use in answering these questions?
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definition of KE, definition of force
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Excellent work on this question.
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`q004. For the system of the preceding, suppose the 30 kg mass is placed on a frictionless horizontal surface, with the 31 kg mass suspended over the pulley. If the system is released from rest and accelerates through a 5 meter displacement:
By how much would its gravitational PE change?
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force applied: 9.8 * 31: 303.8; change in PE: 303.8 * -5: -1519 Joules
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By how much would its KE change?
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1519 joules
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What velocity would it therefore attain?
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1519 = 30.5 * v^2; 49.8 = v^2; 7.06 m/s;
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What would be its acceleration?
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303.8/61 = 4.98 m/s^2
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Assuming this acceleration, what would be the velocity at the end of the 5 meter displacement?
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vf^2 = 9.96 * 5; vf^2 = 49.8 m^2/s^2; vf = 7.05 m/s
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What definitions, laws and procedures did you use in answering these questions?
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definition of KE/PE, change in KE = work done, definition of force
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Excellent work on this question as well.
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You've done a lot of things right and you have just about all the quantities necessary to correct the first three questions.
The last two look great, but the first three need to be revised.
&#Please see my notes and submit a copy of this document with revisions, comments and/or questions, and mark your insertions with &&&& (please mark each insertion at the beginning and at the end).
Be sure to include the entire document, including my notes. &#
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